This batch has passed Schema, deduplication, source-snapshot, and independent-answer recomputation checks; it is still not marked as human-reviewed.
This question bank has corresponding bilingual step-by-step derivations. Source candidate and English ai_candidate states are retained. AI review, finite computation and Lean formal proof are labelled separately.
batch-20261003-r04-quadratic-equation-foundations · 2026-10-03 · 100 questions
This English edition is an AI candidate and does not yet have an OCME Atlas (GPT-6) Twin closure receipt.
Over the reals, is (-2)x²+(-12)x+(-3)=0 a quadratic equation in one variable? Answer Yes or No.
Yes
The quadratic coefficient is a=-2, so the answer is Yes. A quadratic equation requires a≠0, even if a zero-coefficient x² term is written.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-form-validity · candidate
Over the reals, is (-1)x²+(-11)x+(-2)=0 a quadratic equation in one variable? Answer Yes or No.
Yes
The quadratic coefficient is a=-1, so the answer is Yes. A quadratic equation requires a≠0, even if a zero-coefficient x² term is written.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-form-validity · candidate
Over the reals, is (0)x²+(-10)x+(-1)=0 a quadratic equation in one variable? Answer Yes or No.
No
The quadratic coefficient is a=0, so the answer is No. A quadratic equation requires a≠0, even if a zero-coefficient x² term is written.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-form-validity · candidate
Over the reals, is (1)x²+(-9)x+(0)=0 a quadratic equation in one variable? Answer Yes or No.
Yes
The quadratic coefficient is a=1, so the answer is Yes. A quadratic equation requires a≠0, even if a zero-coefficient x² term is written.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-form-validity · candidate
Over the reals, is (2)x²+(-8)x+(1)=0 a quadratic equation in one variable? Answer Yes or No.
Yes
The quadratic coefficient is a=2, so the answer is Yes. A quadratic equation requires a≠0, even if a zero-coefficient x² term is written.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-form-validity · candidate
Over the reals, is (-2)x²+(-7)x+(2)=0 a quadratic equation in one variable? Answer Yes or No.
Yes
The quadratic coefficient is a=-2, so the answer is Yes. A quadratic equation requires a≠0, even if a zero-coefficient x² term is written.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-form-validity · candidate
Over the reals, is (-1)x²+(-6)x+(3)=0 a quadratic equation in one variable? Answer Yes or No.
Yes
The quadratic coefficient is a=-1, so the answer is Yes. A quadratic equation requires a≠0, even if a zero-coefficient x² term is written.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-form-validity · candidate
Over the reals, is (0)x²+(-5)x+(-3)=0 a quadratic equation in one variable? Answer Yes or No.
No
The quadratic coefficient is a=0, so the answer is No. A quadratic equation requires a≠0, even if a zero-coefficient x² term is written.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-form-validity · candidate
Over the reals, is (1)x²+(-4)x+(-2)=0 a quadratic equation in one variable? Answer Yes or No.
Yes
The quadratic coefficient is a=1, so the answer is Yes. A quadratic equation requires a≠0, even if a zero-coefficient x² term is written.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-form-validity · candidate
Over the reals, is (2)x²+(-3)x+(-1)=0 a quadratic equation in one variable? Answer Yes or No.
Yes
The quadratic coefficient is a=2, so the answer is Yes. A quadratic equation requires a≠0, even if a zero-coefficient x² term is written.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-form-validity · candidate
Over the reals, is (-2)x²+(-2)x+(0)=0 a quadratic equation in one variable? Answer Yes or No.
Yes
The quadratic coefficient is a=-2, so the answer is Yes. A quadratic equation requires a≠0, even if a zero-coefficient x² term is written.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-form-validity · candidate
Over the reals, is (-1)x²+(-1)x+(1)=0 a quadratic equation in one variable? Answer Yes or No.
Yes
The quadratic coefficient is a=-1, so the answer is Yes. A quadratic equation requires a≠0, even if a zero-coefficient x² term is written.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-form-validity · candidate
Over the reals, is (0)x²+(0)x+(2)=0 a quadratic equation in one variable? Answer Yes or No.
No
The quadratic coefficient is a=0, so the answer is No. A quadratic equation requires a≠0, even if a zero-coefficient x² term is written.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-form-validity · candidate
Over the reals, is (1)x²+(1)x+(3)=0 a quadratic equation in one variable? Answer Yes or No.
Yes
The quadratic coefficient is a=1, so the answer is Yes. A quadratic equation requires a≠0, even if a zero-coefficient x² term is written.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-form-validity · candidate
Over the reals, is (2)x²+(2)x+(-3)=0 a quadratic equation in one variable? Answer Yes or No.
Yes
The quadratic coefficient is a=2, so the answer is Yes. A quadratic equation requires a≠0, even if a zero-coefficient x² term is written.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-form-validity · candidate
Over the reals, is (-2)x²+(3)x+(-2)=0 a quadratic equation in one variable? Answer Yes or No.
Yes
The quadratic coefficient is a=-2, so the answer is Yes. A quadratic equation requires a≠0, even if a zero-coefficient x² term is written.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-form-validity · candidate
Over the reals, is (-1)x²+(4)x+(-1)=0 a quadratic equation in one variable? Answer Yes or No.
Yes
The quadratic coefficient is a=-1, so the answer is Yes. A quadratic equation requires a≠0, even if a zero-coefficient x² term is written.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-form-validity · candidate
Over the reals, is (0)x²+(5)x+(0)=0 a quadratic equation in one variable? Answer Yes or No.
No
The quadratic coefficient is a=0, so the answer is No. A quadratic equation requires a≠0, even if a zero-coefficient x² term is written.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-form-validity · candidate
Over the reals, is (1)x²+(6)x+(1)=0 a quadratic equation in one variable? Answer Yes or No.
Yes
The quadratic coefficient is a=1, so the answer is Yes. A quadratic equation requires a≠0, even if a zero-coefficient x² term is written.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-form-validity · candidate
Over the reals, is (2)x²+(7)x+(2)=0 a quadratic equation in one variable? Answer Yes or No.
Yes
The quadratic coefficient is a=2, so the answer is Yes. A quadratic equation requires a≠0, even if a zero-coefficient x² term is written.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-form-validity · candidate
Over the reals, is (-2)x²+(8)x+(3)=0 a quadratic equation in one variable? Answer Yes or No.
Yes
The quadratic coefficient is a=-2, so the answer is Yes. A quadratic equation requires a≠0, even if a zero-coefficient x² term is written.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-form-validity · candidate
Over the reals, is (-1)x²+(9)x+(-3)=0 a quadratic equation in one variable? Answer Yes or No.
Yes
The quadratic coefficient is a=-1, so the answer is Yes. A quadratic equation requires a≠0, even if a zero-coefficient x² term is written.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-form-validity · candidate
Over the reals, is (0)x²+(10)x+(-2)=0 a quadratic equation in one variable? Answer Yes or No.
No
The quadratic coefficient is a=0, so the answer is No. A quadratic equation requires a≠0, even if a zero-coefficient x² term is written.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-form-validity · candidate
Over the reals, is (1)x²+(11)x+(-1)=0 a quadratic equation in one variable? Answer Yes or No.
Yes
The quadratic coefficient is a=1, so the answer is Yes. A quadratic equation requires a≠0, even if a zero-coefficient x² term is written.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-form-validity · candidate
Over the reals, is (2)x²+(12)x+(0)=0 a quadratic equation in one variable? Answer Yes or No.
Yes
The quadratic coefficient is a=2, so the answer is Yes. A quadratic equation requires a≠0, even if a zero-coefficient x² term is written.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-form-validity · candidate
Over the reals, (1)x²+(0)x+(1)=0 has a≠0. Find the discriminant D and the number of distinct real roots.
D=-4; distinct real roots=0
D=-4; distinct real roots=0. Compute D=b²−4ac; negative, zero, and positive D give zero, one, and two distinct real roots respectively. A repeated root is counted once.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-discriminant · candidate
Over the reals, (1)x²+(0)x+(0)=0 has a≠0. Find the discriminant D and the number of distinct real roots.
D=0; distinct real roots=1
D=0; distinct real roots=1. Compute D=b²−4ac; negative, zero, and positive D give zero, one, and two distinct real roots respectively. A repeated root is counted once.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-discriminant · candidate
Over the reals, (-1)x²+(0)x+(1)=0 has a≠0. Find the discriminant D and the number of distinct real roots.
D=4; distinct real roots=2
D=4; distinct real roots=2. Compute D=b²−4ac; negative, zero, and positive D give zero, one, and two distinct real roots respectively. A repeated root is counted once.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-discriminant · candidate
Over the reals, (1)x²+(2)x+(2)=0 has a≠0. Find the discriminant D and the number of distinct real roots.
D=-4; distinct real roots=0
D=-4; distinct real roots=0. Compute D=b²−4ac; negative, zero, and positive D give zero, one, and two distinct real roots respectively. A repeated root is counted once.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-discriminant · candidate
Over the reals, (1)x²+(2)x+(1)=0 has a≠0. Find the discriminant D and the number of distinct real roots.
D=0; distinct real roots=1
D=0; distinct real roots=1. Compute D=b²−4ac; negative, zero, and positive D give zero, one, and two distinct real roots respectively. A repeated root is counted once.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-discriminant · candidate
Over the reals, (-1)x²+(1)x+(2)=0 has a≠0. Find the discriminant D and the number of distinct real roots.
D=9; distinct real roots=2
D=9; distinct real roots=2. Compute D=b²−4ac; negative, zero, and positive D give zero, one, and two distinct real roots respectively. A repeated root is counted once.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-discriminant · candidate
Over the reals, (1)x²+(4)x+(5)=0 has a≠0. Find the discriminant D and the number of distinct real roots.
D=-4; distinct real roots=0
D=-4; distinct real roots=0. Compute D=b²−4ac; negative, zero, and positive D give zero, one, and two distinct real roots respectively. A repeated root is counted once.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-discriminant · candidate
Over the reals, (1)x²+(4)x+(4)=0 has a≠0. Find the discriminant D and the number of distinct real roots.
D=0; distinct real roots=1
D=0; distinct real roots=1. Compute D=b²−4ac; negative, zero, and positive D give zero, one, and two distinct real roots respectively. A repeated root is counted once.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-discriminant · candidate
Over the reals, (-1)x²+(2)x+(3)=0 has a≠0. Find the discriminant D and the number of distinct real roots.
D=16; distinct real roots=2
D=16; distinct real roots=2. Compute D=b²−4ac; negative, zero, and positive D give zero, one, and two distinct real roots respectively. A repeated root is counted once.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-discriminant · candidate
Over the reals, (1)x²+(6)x+(10)=0 has a≠0. Find the discriminant D and the number of distinct real roots.
D=-4; distinct real roots=0
D=-4; distinct real roots=0. Compute D=b²−4ac; negative, zero, and positive D give zero, one, and two distinct real roots respectively. A repeated root is counted once.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-discriminant · candidate
Over the reals, (1)x²+(6)x+(9)=0 has a≠0. Find the discriminant D and the number of distinct real roots.
D=0; distinct real roots=1
D=0; distinct real roots=1. Compute D=b²−4ac; negative, zero, and positive D give zero, one, and two distinct real roots respectively. A repeated root is counted once.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-discriminant · candidate
Over the reals, (-1)x²+(3)x+(4)=0 has a≠0. Find the discriminant D and the number of distinct real roots.
D=25; distinct real roots=2
D=25; distinct real roots=2. Compute D=b²−4ac; negative, zero, and positive D give zero, one, and two distinct real roots respectively. A repeated root is counted once.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-discriminant · candidate
Over the reals, (1)x²+(8)x+(17)=0 has a≠0. Find the discriminant D and the number of distinct real roots.
D=-4; distinct real roots=0
D=-4; distinct real roots=0. Compute D=b²−4ac; negative, zero, and positive D give zero, one, and two distinct real roots respectively. A repeated root is counted once.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-discriminant · candidate
Over the reals, (1)x²+(8)x+(16)=0 has a≠0. Find the discriminant D and the number of distinct real roots.
D=0; distinct real roots=1
D=0; distinct real roots=1. Compute D=b²−4ac; negative, zero, and positive D give zero, one, and two distinct real roots respectively. A repeated root is counted once.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-discriminant · candidate
Over the reals, (-1)x²+(4)x+(5)=0 has a≠0. Find the discriminant D and the number of distinct real roots.
D=36; distinct real roots=2
D=36; distinct real roots=2. Compute D=b²−4ac; negative, zero, and positive D give zero, one, and two distinct real roots respectively. A repeated root is counted once.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-discriminant · candidate
Over the reals, (1)x²+(10)x+(26)=0 has a≠0. Find the discriminant D and the number of distinct real roots.
D=-4; distinct real roots=0
D=-4; distinct real roots=0. Compute D=b²−4ac; negative, zero, and positive D give zero, one, and two distinct real roots respectively. A repeated root is counted once.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-discriminant · candidate
Over the reals, (1)x²+(10)x+(25)=0 has a≠0. Find the discriminant D and the number of distinct real roots.
D=0; distinct real roots=1
D=0; distinct real roots=1. Compute D=b²−4ac; negative, zero, and positive D give zero, one, and two distinct real roots respectively. A repeated root is counted once.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-discriminant · candidate
Over the reals, (-1)x²+(5)x+(6)=0 has a≠0. Find the discriminant D and the number of distinct real roots.
D=49; distinct real roots=2
D=49; distinct real roots=2. Compute D=b²−4ac; negative, zero, and positive D give zero, one, and two distinct real roots respectively. A repeated root is counted once.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-discriminant · candidate
Over the reals, (1)x²+(12)x+(37)=0 has a≠0. Find the discriminant D and the number of distinct real roots.
D=-4; distinct real roots=0
D=-4; distinct real roots=0. Compute D=b²−4ac; negative, zero, and positive D give zero, one, and two distinct real roots respectively. A repeated root is counted once.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-discriminant · candidate
Over the reals, (1)x²+(12)x+(36)=0 has a≠0. Find the discriminant D and the number of distinct real roots.
D=0; distinct real roots=1
D=0; distinct real roots=1. Compute D=b²−4ac; negative, zero, and positive D give zero, one, and two distinct real roots respectively. A repeated root is counted once.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-discriminant · candidate
Over the reals, (-1)x²+(6)x+(7)=0 has a≠0. Find the discriminant D and the number of distinct real roots.
D=64; distinct real roots=2
D=64; distinct real roots=2. Compute D=b²−4ac; negative, zero, and positive D give zero, one, and two distinct real roots respectively. A repeated root is counted once.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-discriminant · candidate
Over the reals, (1)x²+(14)x+(50)=0 has a≠0. Find the discriminant D and the number of distinct real roots.
D=-4; distinct real roots=0
D=-4; distinct real roots=0. Compute D=b²−4ac; negative, zero, and positive D give zero, one, and two distinct real roots respectively. A repeated root is counted once.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-discriminant · candidate
Over the reals, (1)x²+(14)x+(49)=0 has a≠0. Find the discriminant D and the number of distinct real roots.
D=0; distinct real roots=1
D=0; distinct real roots=1. Compute D=b²−4ac; negative, zero, and positive D give zero, one, and two distinct real roots respectively. A repeated root is counted once.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-discriminant · candidate
Over the reals, (-1)x²+(7)x+(8)=0 has a≠0. Find the discriminant D and the number of distinct real roots.
D=81; distinct real roots=2
D=81; distinct real roots=2. Compute D=b²−4ac; negative, zero, and positive D give zero, one, and two distinct real roots respectively. A repeated root is counted once.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-discriminant · candidate
Over the reals, (1)x²+(16)x+(65)=0 has a≠0. Find the discriminant D and the number of distinct real roots.
D=-4; distinct real roots=0
D=-4; distinct real roots=0. Compute D=b²−4ac; negative, zero, and positive D give zero, one, and two distinct real roots respectively. A repeated root is counted once.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-discriminant · candidate
Over the reals, (2)x²+(16)x+(24)=0 has a≠0. Is the reduced rational candidate x=-6/1, with positive denominator, a solution?
Yes
Since q=1>0, clear denominators by multiplying the substituted expression by q². Its numerator is 0; comparison with zero gives Yes.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-root-verification · candidate
Over the reals, (-4)x²+(-26)x+(-22)=0 has a≠0. Is the reduced rational candidate x=-5/1, with positive denominator, a solution?
No
Since q=1>0, clear denominators by multiplying the substituted expression by q². Its numerator is 8; comparison with zero gives No.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-root-verification · candidate
Over the reals, (6)x²+(30)x+(0)=0 has a≠0. Is the reduced rational candidate x=-5/1, with positive denominator, a solution?
Yes
Since q=1>0, clear denominators by multiplying the substituted expression by q². Its numerator is 0; comparison with zero gives Yes.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-root-verification · candidate
Over the reals, (-2)x²+(-7)x+(9)=0 has a≠0. Is the reduced rational candidate x=-4/1, with positive denominator, a solution?
No
Since q=1>0, clear denominators by multiplying the substituted expression by q². Its numerator is 5; comparison with zero gives No.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-root-verification · candidate
Over the reals, (4)x²+(8)x+(-32)=0 has a≠0. Is the reduced rational candidate x=-4/1, with positive denominator, a solution?
Yes
Since q=1>0, clear denominators by multiplying the substituted expression by q². Its numerator is 0; comparison with zero gives Yes.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-root-verification · candidate
Over the reals, (-6)x²+(-33)x+(-42)=0 has a≠0. Is the reduced rational candidate x=-3/1, with positive denominator, a solution?
No
Since q=1>0, clear denominators by multiplying the substituted expression by q². Its numerator is 3; comparison with zero gives No.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-root-verification · candidate
Over the reals, (2)x²+(8)x+(6)=0 has a≠0. Is the reduced rational candidate x=-3/1, with positive denominator, a solution?
Yes
Since q=1>0, clear denominators by multiplying the substituted expression by q². Its numerator is 0; comparison with zero gives Yes.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-root-verification · candidate
Over the reals, (-4)x²+(-10)x+(0)=0 has a≠0. Is the reduced rational candidate x=-2/1, with positive denominator, a solution?
No
Since q=1>0, clear denominators by multiplying the substituted expression by q². Its numerator is 4; comparison with zero gives No.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-root-verification · candidate
Over the reals, (6)x²+(6)x+(-12)=0 has a≠0. Is the reduced rational candidate x=-2/1, with positive denominator, a solution?
Yes
Since q=1>0, clear denominators by multiplying the substituted expression by q². Its numerator is 0; comparison with zero gives Yes.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-root-verification · candidate
Over the reals, (-2)x²+(1)x+(6)=0 has a≠0. Is the reduced rational candidate x=-1/1, with positive denominator, a solution?
No
Since q=1>0, clear denominators by multiplying the substituted expression by q². Its numerator is 3; comparison with zero gives No.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-root-verification · candidate
Over the reals, (4)x²+(12)x+(8)=0 has a≠0. Is the reduced rational candidate x=-1/1, with positive denominator, a solution?
Yes
Since q=1>0, clear denominators by multiplying the substituted expression by q². Its numerator is 0; comparison with zero gives Yes.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-root-verification · candidate
Over the reals, (-6)x²+(-9)x+(-3)=0 has a≠0. Is the reduced rational candidate x=0/1, with positive denominator, a solution?
No
Since q=1>0, clear denominators by multiplying the substituted expression by q². Its numerator is -3; comparison with zero gives No.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-root-verification · candidate
Over the reals, (2)x²+(0)x+(0)=0 has a≠0. Is the reduced rational candidate x=0/1, with positive denominator, a solution?
Yes
Since q=1>0, clear denominators by multiplying the substituted expression by q². Its numerator is 0; comparison with zero gives Yes.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-root-verification · candidate
Over the reals, (-4)x²+(6)x+(-2)=0 has a≠0. Is the reduced rational candidate x=1/1, with positive denominator, a solution?
Yes
Since q=1>0, clear denominators by multiplying the substituted expression by q². Its numerator is 0; comparison with zero gives Yes.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-root-verification · candidate
Over the reals, (6)x²+(-18)x+(12)=0 has a≠0. Is the reduced rational candidate x=1/1, with positive denominator, a solution?
Yes
Since q=1>0, clear denominators by multiplying the substituted expression by q². Its numerator is 0; comparison with zero gives Yes.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-root-verification · candidate
Over the reals, (-2)x²+(-1)x+(6)=0 has a≠0. Is the reduced rational candidate x=2/1, with positive denominator, a solution?
No
Since q=1>0, clear denominators by multiplying the substituted expression by q². Its numerator is -4; comparison with zero gives No.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-root-verification · candidate
Over the reals, (4)x²+(-4)x+(-8)=0 has a≠0. Is the reduced rational candidate x=2/1, with positive denominator, a solution?
Yes
Since q=1>0, clear denominators by multiplying the substituted expression by q². Its numerator is 0; comparison with zero gives Yes.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-root-verification · candidate
Over the reals, (-6)x²+(15)x+(0)=0 has a≠0. Is the reduced rational candidate x=3/1, with positive denominator, a solution?
No
Since q=1>0, clear denominators by multiplying the substituted expression by q². Its numerator is -9; comparison with zero gives No.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-root-verification · candidate
Over the reals, (2)x²+(-8)x+(6)=0 has a≠0. Is the reduced rational candidate x=3/1, with positive denominator, a solution?
Yes
Since q=1>0, clear denominators by multiplying the substituted expression by q². Its numerator is 0; comparison with zero gives Yes.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-root-verification · candidate
Over the reals, (-4)x²+(22)x+(-28)=0 has a≠0. Is the reduced rational candidate x=4/1, with positive denominator, a solution?
No
Since q=1>0, clear denominators by multiplying the substituted expression by q². Its numerator is -4; comparison with zero gives No.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-root-verification · candidate
Over the reals, (6)x²+(-12)x+(-48)=0 has a≠0. Is the reduced rational candidate x=4/1, with positive denominator, a solution?
Yes
Since q=1>0, clear denominators by multiplying the substituted expression by q². Its numerator is 0; comparison with zero gives Yes.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-root-verification · candidate
Over the reals, (-2)x²+(7)x+(9)=0 has a≠0. Is the reduced rational candidate x=5/1, with positive denominator, a solution?
No
Since q=1>0, clear denominators by multiplying the substituted expression by q². Its numerator is -6; comparison with zero gives No.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-root-verification · candidate
Over the reals, (4)x²+(-20)x+(0)=0 has a≠0. Is the reduced rational candidate x=5/1, with positive denominator, a solution?
Yes
Since q=1>0, clear denominators by multiplying the substituted expression by q². Its numerator is 0; comparison with zero gives Yes.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-root-verification · candidate
Over the reals, (-6)x²+(39)x+(-33)=0 has a≠0. Is the reduced rational candidate x=6/1, with positive denominator, a solution?
No
Since q=1>0, clear denominators by multiplying the substituted expression by q². Its numerator is -15; comparison with zero gives No.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-root-verification · candidate
Over the reals, (2)x²+(-16)x+(24)=0 has a≠0. Is the reduced rational candidate x=6/1, with positive denominator, a solution?
Yes
Since q=1>0, clear denominators by multiplying the substituted expression by q². Its numerator is 0; comparison with zero gives Yes.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-root-verification · candidate
Over the reals, (1)x²+(0)x+(0)=0 has a≠0 and a nonnegative perfect-square discriminant. List the exact root set in increasing order as reduced fractions with positive denominators, writing each repeated value once.
{0/1}
D=0. Apply (-b±√D)/(2a), reduce the rational roots, sort them, and remove duplicate values to obtain {0/1}. The comparison is exact.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-exact-roots · candidate
Over the reals, (-1)x²+(2)x+(-1)=0 has a≠0 and a nonnegative perfect-square discriminant. List the exact root set in increasing order as reduced fractions with positive denominators, writing each repeated value once.
{1/1}
D=0. Apply (-b±√D)/(2a), reduce the rational roots, sort them, and remove duplicate values to obtain {1/1}. The comparison is exact.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-exact-roots · candidate
Over the reals, (6)x²+(32)x+(10)=0 has a≠0 and a nonnegative perfect-square discriminant. List the exact root set in increasing order as reduced fractions with positive denominators, writing each repeated value once.
{-5/1, -1/3}
D=784. Apply (-b±√D)/(2a), reduce the rational roots, sort them, and remove duplicate values to obtain {-5/1, -1/3}. The comparison is exact.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-exact-roots · candidate
Over the reals, (-6)x²+(-27)x+(0)=0 has a≠0 and a nonnegative perfect-square discriminant. List the exact root set in increasing order as reduced fractions with positive denominators, writing each repeated value once.
{-9/2, 0/1}
D=729. Apply (-b±√D)/(2a), reduce the rational roots, sort them, and remove duplicate values to obtain {-9/2, 0/1}. The comparison is exact.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-exact-roots · candidate
Over the reals, (6)x²+(22)x+(-8)=0 has a≠0 and a nonnegative perfect-square discriminant. List the exact root set in increasing order as reduced fractions with positive denominators, writing each repeated value once.
{-4/1, 1/3}
D=676. Apply (-b±√D)/(2a), reduce the rational roots, sort them, and remove duplicate values to obtain {-4/1, 1/3}. The comparison is exact.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-exact-roots · candidate
Over the reals, (-6)x²+(-17)x+(14)=0 has a≠0 and a nonnegative perfect-square discriminant. List the exact root set in increasing order as reduced fractions with positive denominators, writing each repeated value once.
{-7/2, 2/3}
D=625. Apply (-b±√D)/(2a), reduce the rational roots, sort them, and remove duplicate values to obtain {-7/2, 2/3}. The comparison is exact.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-exact-roots · candidate
Over the reals, (6)x²+(12)x+(-18)=0 has a≠0 and a nonnegative perfect-square discriminant. List the exact root set in increasing order as reduced fractions with positive denominators, writing each repeated value once.
{-3/1, 1/1}
D=576. Apply (-b±√D)/(2a), reduce the rational roots, sort them, and remove duplicate values to obtain {-3/1, 1/1}. The comparison is exact.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-exact-roots · candidate
Over the reals, (-6)x²+(-21)x+(-15)=0 has a≠0 and a nonnegative perfect-square discriminant. List the exact root set in increasing order as reduced fractions with positive denominators, writing each repeated value once.
{-5/2, -1/1}
D=81. Apply (-b±√D)/(2a), reduce the rational roots, sort them, and remove duplicate values to obtain {-5/2, -1/1}. The comparison is exact.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-exact-roots · candidate
Over the reals, (6)x²+(16)x+(8)=0 has a≠0 and a nonnegative perfect-square discriminant. List the exact root set in increasing order as reduced fractions with positive denominators, writing each repeated value once.
{-2/1, -2/3}
D=64. Apply (-b±√D)/(2a), reduce the rational roots, sort them, and remove duplicate values to obtain {-2/1, -2/3}. The comparison is exact.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-exact-roots · candidate
Over the reals, (-6)x²+(-11)x+(-3)=0 has a≠0 and a nonnegative perfect-square discriminant. List the exact root set in increasing order as reduced fractions with positive denominators, writing each repeated value once.
{-3/2, -1/3}
D=49. Apply (-b±√D)/(2a), reduce the rational roots, sort them, and remove duplicate values to obtain {-3/2, -1/3}. The comparison is exact.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-exact-roots · candidate
Over the reals, (6)x²+(6)x+(0)=0 has a≠0 and a nonnegative perfect-square discriminant. List the exact root set in increasing order as reduced fractions with positive denominators, writing each repeated value once.
{-1/1, 0/1}
D=36. Apply (-b±√D)/(2a), reduce the rational roots, sort them, and remove duplicate values to obtain {-1/1, 0/1}. The comparison is exact.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-exact-roots · candidate
Over the reals, (-6)x²+(-1)x+(1)=0 has a≠0 and a nonnegative perfect-square discriminant. List the exact root set in increasing order as reduced fractions with positive denominators, writing each repeated value once.
{-1/2, 1/3}
D=25. Apply (-b±√D)/(2a), reduce the rational roots, sort them, and remove duplicate values to obtain {-1/2, 1/3}. The comparison is exact.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-exact-roots · candidate
Over the reals, (6)x²+(-4)x+(0)=0 has a≠0 and a nonnegative perfect-square discriminant. List the exact root set in increasing order as reduced fractions with positive denominators, writing each repeated value once.
{0/1, 2/3}
D=16. Apply (-b±√D)/(2a), reduce the rational roots, sort them, and remove duplicate values to obtain {0/1, 2/3}. The comparison is exact.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-exact-roots · candidate
Over the reals, (-6)x²+(9)x+(-3)=0 has a≠0 and a nonnegative perfect-square discriminant. List the exact root set in increasing order as reduced fractions with positive denominators, writing each repeated value once.
{1/2, 1/1}
D=9. Apply (-b±√D)/(2a), reduce the rational roots, sort them, and remove duplicate values to obtain {1/2, 1/1}. The comparison is exact.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-exact-roots · candidate
Over the reals, (6)x²+(0)x+(-6)=0 has a≠0 and a nonnegative perfect-square discriminant. List the exact root set in increasing order as reduced fractions with positive denominators, writing each repeated value once.
{-1/1, 1/1}
D=144. Apply (-b±√D)/(2a), reduce the rational roots, sort them, and remove duplicate values to obtain {-1/1, 1/1}. The comparison is exact.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-exact-roots · candidate
Over the reals, (-6)x²+(5)x+(6)=0 has a≠0 and a nonnegative perfect-square discriminant. List the exact root set in increasing order as reduced fractions with positive denominators, writing each repeated value once.
{-2/3, 3/2}
D=169. Apply (-b±√D)/(2a), reduce the rational roots, sort them, and remove duplicate values to obtain {-2/3, 3/2}. The comparison is exact.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-exact-roots · candidate
Over the reals, (6)x²+(-10)x+(-4)=0 has a≠0 and a nonnegative perfect-square discriminant. List the exact root set in increasing order as reduced fractions with positive denominators, writing each repeated value once.
{-1/3, 2/1}
D=196. Apply (-b±√D)/(2a), reduce the rational roots, sort them, and remove duplicate values to obtain {-1/3, 2/1}. The comparison is exact.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-exact-roots · candidate
Over the reals, (-6)x²+(15)x+(0)=0 has a≠0 and a nonnegative perfect-square discriminant. List the exact root set in increasing order as reduced fractions with positive denominators, writing each repeated value once.
{0/1, 5/2}
D=225. Apply (-b±√D)/(2a), reduce the rational roots, sort them, and remove duplicate values to obtain {0/1, 5/2}. The comparison is exact.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-exact-roots · candidate
Over the reals, (6)x²+(-20)x+(6)=0 has a≠0 and a nonnegative perfect-square discriminant. List the exact root set in increasing order as reduced fractions with positive denominators, writing each repeated value once.
{1/3, 3/1}
D=256. Apply (-b±√D)/(2a), reduce the rational roots, sort them, and remove duplicate values to obtain {1/3, 3/1}. The comparison is exact.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-exact-roots · candidate
Over the reals, (-6)x²+(25)x+(-14)=0 has a≠0 and a nonnegative perfect-square discriminant. List the exact root set in increasing order as reduced fractions with positive denominators, writing each repeated value once.
{2/3, 7/2}
D=289. Apply (-b±√D)/(2a), reduce the rational roots, sort them, and remove duplicate values to obtain {2/3, 7/2}. The comparison is exact.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-exact-roots · candidate
Over the reals, (6)x²+(-30)x+(24)=0 has a≠0 and a nonnegative perfect-square discriminant. List the exact root set in increasing order as reduced fractions with positive denominators, writing each repeated value once.
{1/1, 4/1}
D=324. Apply (-b±√D)/(2a), reduce the rational roots, sort them, and remove duplicate values to obtain {1/1, 4/1}. The comparison is exact.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-exact-roots · candidate
Over the reals, (-6)x²+(21)x+(27)=0 has a≠0 and a nonnegative perfect-square discriminant. List the exact root set in increasing order as reduced fractions with positive denominators, writing each repeated value once.
{-1/1, 9/2}
D=1089. Apply (-b±√D)/(2a), reduce the rational roots, sort them, and remove duplicate values to obtain {-1/1, 9/2}. The comparison is exact.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-exact-roots · candidate
Over the reals, (6)x²+(-26)x+(-20)=0 has a≠0 and a nonnegative perfect-square discriminant. List the exact root set in increasing order as reduced fractions with positive denominators, writing each repeated value once.
{-2/3, 5/1}
D=1156. Apply (-b±√D)/(2a), reduce the rational roots, sort them, and remove duplicate values to obtain {-2/3, 5/1}. The comparison is exact.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-exact-roots · candidate
Over the reals, (-6)x²+(31)x+(11)=0 has a≠0 and a nonnegative perfect-square discriminant. List the exact root set in increasing order as reduced fractions with positive denominators, writing each repeated value once.
{-1/3, 11/2}
D=1225. Apply (-b±√D)/(2a), reduce the rational roots, sort them, and remove duplicate values to obtain {-1/3, 11/2}. The comparison is exact.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-exact-roots · candidate
Over the reals, (6)x²+(-36)x+(0)=0 has a≠0 and a nonnegative perfect-square discriminant. List the exact root set in increasing order as reduced fractions with positive denominators, writing each repeated value once.
{0/1, 6/1}
D=1296. Apply (-b±√D)/(2a), reduce the rational roots, sort them, and remove duplicate values to obtain {0/1, 6/1}. The comparison is exact.
Reviewed step-by-step derivation
Quadratic equations over the reals · qf-quadratic-exact-roots · candidate
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