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While rewriting a quadratic equation into the form ax^2 + bx + c = 0
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While rewriting a quadratic equation into the form \(ax^2 + bx + c = 0\), John and Katie both made errors. John made an error that resulted in c being incorrect while a and b were correct. Katie made an error that resulted in b being incorrect while a and c were correct. If John obtained 6 and 2 as solutions and Katie obtained -7 and -1 as solutions, and neither of them made any additional errors, what were the correct solutions to the original quadratic equation?

A. -8 and 7
B. -8 and 12
C. -6 and -2
D. 1 and 6
E. 1 and 7
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While rewriting a quadratic equation into the form ax^2 + bx + c = 0
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05 Aug 2023, 07:19
Bunuel
While rewriting a quadratic equation into the form \(ax^2 + bx + c = 0\), John and Katie both made errors. John made an error that resulted in c being incorrect while a and b were correct. Katie made an error that resulted in b being incorrect while a and c were correct. If John obtained 6 and 2 as solutions and Katie obtained -7 and -1 as solutions, and neither of them made any additional errors, what were the correct solutions to the original quadratic equation?

A. -8 and 7
B. -8 and 12
C. -6 and -2
D. 1 and 6
E. 1 and 7
Show more

Solution obtained by John

\((x-6)(x-2) = 0\)

\(x^2 -8x + 12 = 0\)

We know that the value of +12 is incorrect.

Solution obtained by Katie

\((x+7)(x+1) = 0\)

\(x^2 + 8x + 7 = 0\)

From the question stem we know that the value +7x is incorrect.

Combined

\(x^2 -8x + 7 = 0\)

\(x^2 -7x - x + 7 = 0\)

\(x(x -7) -1 (x - 7) = 0\)

\((x-1)(x-7) = 0\)

Therefore the solutions are x = 1 , and x = 7

Option E

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While rewriting a quadratic equation into the form ax^2 + bx + c = 0
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07 Aug 2023, 06:25
Bunuel
While rewriting a quadratic equation into the form \(ax^2 + bx + c = 0\), John and Katie both made errors. John made an error that resulted in c being incorrect while a and b were correct. Katie made an error that resulted in b being incorrect while a and c were correct. If John obtained 6 and 2 as solutions and Katie obtained -7 and -1 as solutions, and neither of them made any additional errors, what were the correct solutions to the original quadratic equation?

A. -8 and 7
B. -8 and 12
C. -6 and -2
D. 1 and 6
E. 1 and 7
Show more

John’s incorrectly written equation was:
(x – 6)(x – 2) = 0
x^2 – 8x + 12 = 0

Since a and b were correct in John’s equation, we have:
a = 1 and b = -8

Katie’s incorrectly written equation was:
(x + 7)(x + 1) = 0
x^2 + 8x + 7 = 0

Since a and c were correct in Katie’s equation, we have:
a = 1 and c = 7

So, in the original equation the coefficients are:
a = 1, b = -8, and c = 7

The correctly written original equation and its solutions are:
x^2 – 8b + 7 = 0
(x – 1)(x – 7) = 0 => x = 1, x = 7

Answer: E

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While rewriting a quadratic equation into the form ax^2 + bx + c = 0
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06 Aug 2023, 22:54
While rewriting a quadratic equation into the form ax2+bx+c=0, John and Katie both made errors. John made an error that resulted in c being incorrect while a and b were correct. Katie made an error that resulted in b being incorrect while a and c were correct. If John obtained 6 and 2 as solutions and Katie obtained -7 and -1 as solutions, and neither of them made any additional errors, what were the correct solutions to the original quadratic equation?

We know that if ax2+bx+c=0
Product of roots = c/a and sum of roots = -b/a

In the case of John, c being incorrect means the product of roots is incorrect but the sum is correct
Hence the sum of roots for the correct equation is 6+2=8

In the case of Katie, b being incorrect means the sum of roots is incorrect but the product is correct
Hence the product of roots for the correct equation is 7

therefore the correct equation is

=>x^2-8x+7=0
=>x^2-7x-x+7=0
=>x(x-7)-(x-7)=0
=>(x-1)(x-7)=0

hence the root of correct equations are 1 & 7

Hence E
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While rewriting a quadratic equation into the form ax^2 + bx + c = 0
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24 Feb 2024, 20:59
Not sure if this is logically correct:

­The vertex（middle point) of a Quadratic equation of one variable

equals to 

 -b/2a，[（4ac-b 2）/4a]

Since John get a and b correct, means the x axis value of John is the same as the corrected function

Meanwhile, c means the intercept cordinate of the function to the y axis when x equals to 0,
While Kaite get c correct, means the y axis value for vertex（middle point) is the same as the corrected function

So just simply move Katie function drawing to the corrected X axis value,
and simply added number from the middle point: 4+3 =7; 4 -3 = 1­
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While rewriting a quadratic equation into the form ax^2 + bx + c = 0
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21 Apr 2024, 07:58
­Gmatophobia, your explanation was so helpful! Can you (or someone) please explain to me how you got from 6 and 2 to (x-6)(x-2)? I would have written it as (x+6)(x+2) since the solution was positive (and vice versa for Katie's solutions). Thank you in advance!
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While rewriting a quadratic equation into the form ax^2 + bx + c = 0
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21 Apr 2024, 08:11
 
josiewhoasie
­Gmatophobia, your explanation was so helpful! Can you (or someone) please explain to me how you got from 6 and 2 to (x-6)(x-2)? I would have written it as (x+6)(x+2) since the solution was positive (and vice versa for Katie's solutions). Thank you in advance!
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The solutions of (x - 6)(x - 2) = 0 are x = 6 and x = 2. This is because for (x - 6)(x - 2) = 0 to hold, one of the factors must be zero: x - 6 = 0 implies x = 6, and x - 2 = 0 implies x = 2. A similar logic applies to (x + 6)(x + 2) = 0, where the solutions are x = -6 and x = -2 because x + 6 = 0 yields x = -6, and x + 2 = 0 yields x = -2.

Hope it helps. 

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While rewriting a quadratic equation into the form ax^2 + bx + c = 0
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12 May 2024, 04:36
­Standard quadratic business:


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While rewriting a quadratic equation into the form ax^2 + bx + c = 0
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12 May 2024, 15:50
Thank you both so much! This helped a lot, really appreciate you both.
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While rewriting a quadratic equation into the form ax^2 + bx + c = 0
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31 May 2024, 21:16
Is it possible to use the following easy approach? 
(x + q)(x + p) = 0
q * p = c
q + p = b

If Katie's (-7;-1) solution is correct for c -> so c = 7
That means ONLY [(-7) * (-1) = 7] OR [7 * 1 = 7] are the possible solutions

The only answer that has 7 and 1 is E, correct?­
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While rewriting a quadratic equation into the form ax^2 + bx + c = 0
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03 Aug 2024, 04:22
Bunuel
While rewriting a quadratic equation into the form \(ax^2 + bx + c = 0\), John and Katie both made errors. John made an error that resulted in c being incorrect while a and b were correct. Katie made an error that resulted in b being incorrect while a and c were correct. If John obtained 6 and 2 as solutions and Katie obtained -7 and -1 as solutions, and neither of them made any additional errors, what were the correct solutions to the original quadratic equation?

A. -8 and 7
B. -8 and 12
C. -6 and -2
D. 1 and 6
E. 1 and 7
Show more
­
­A SIMPLE APPROACH
for any quadratic equation \(ax^2 + bx + c = 0\), there are two roots α and β
Also, the Sum of the roots, α + β = -b/a and
Also, the Product of the roots, α * β = c/a

Since, John got a and b correct therefore his sum of the roots (-b/a) should be correct [Roots as per John are 6 and 2)
i.e. Sum of the roots = 6+2 = 8 (CORRECT)

STOP-1: P.S. Only Option E Satisfies (Sum of roots 8) hence the correct Answer.

Let's proceed anyway

Since, Katie got a and c correct therefore her Product of the roots (c/a) should be correct [Roots as per Katie are -7 and -1)
i.e. Product of the roots = (-7)*(-1) = 7 (CORRECT)

STOP-2: P.S. Only Option E Satisfies (product of roots 7) hence the correct Answer.

Let's proceed anyway

Now, we just have to think of two numbers (roots) whose sum is 8 and the product is 7

Now we know that those two numbers are 7 and 1

Answer: Option E­

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While rewriting a quadratic equation into the form ax^2 + bx + c = 0
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06 Nov 2024, 14:52
For a quadratic equation, ax^2 + bx + c = 0

Sum of the roots = -b/a

product of the roots = c/a

John has considered correct values of a and b, hence the sum of roots he got must be equal to sum of correct roots = 6+2 = 8

Katie has considered correct values of a and c, hence the product of roots he got must be equal to product of correct roots = (-7)(-1) = 7

From the given options, E : 1, 7 satisfy the condition that sum = 8 and product = 7.

Hence, E is the correct answer.

Bunuel
While rewriting a quadratic equation into the form \(ax^2 + bx + c = 0\), John and Katie both made errors. John made an error that resulted in c being incorrect while a and b were correct. Katie made an error that resulted in b being incorrect while a and c were correct. If John obtained 6 and 2 as solutions and Katie obtained -7 and -1 as solutions, and neither of them made any additional errors, what were the correct solutions to the original quadratic equation?

A. -8 and 7
B. -8 and 12
C. -6 and -2
D. 1 and 6
E. 1 and 7
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While rewriting a quadratic equation into the form ax^2 + bx + c = 0
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04 Feb 2025, 18:20
Solution obtained by Katie (x+7)(x+1)=0
should be x^{2}+8x+7=0 Right ?
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Bunuel
While rewriting a quadratic equation into the form \(ax^2 + bx + c = 0\), John and Katie both made errors. John made an error that resulted in c being incorrect while a and b were correct. Katie made an error that resulted in b being incorrect while a and c were correct. If John obtained 6 and 2 as solutions and Katie obtained -7 and -1 as solutions, and neither of them made any additional errors, what were the correct solutions to the original quadratic equation?

A. -8 and 7
B. -8 and 12
C. -6 and -2
D. 1 and 6
E. 1 and 7

Solution obtained by John

\((x-6)(x-2) = 0\)

\(x^2 -8x + 12 = 0\)

We know that the value of +12 is incorrect.

Solution obtained by Katie

\((x+7)(x+1) = 0\)

\(x^2 + 7x + 7 = 0\)

From the question stem we know that the value +7x is incorrect.

Combined

\(x^2 -8x + 7 = 0\)

\(x^2 -7x - x + 7 = 0\)

\(x(x -7) -1 (x - 7) = 0\)

\((x-1)(x-7) = 0\)

Therefore the solutions are x = 1 , and x = 7

Option E
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While rewriting a quadratic equation into the form ax^2 + bx + c = 0
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15 May 2025, 03:45
is there a reason why we don't consider the option that in the quadratic equation, ax^2 + bx + c = 0 'a' could be greater than 1?
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While rewriting a quadratic equation into the form ax^2 + bx + c = 0
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15 May 2025, 06:12
sophia1234
While rewriting a quadratic equation into the form \(ax^2 + bx + c = 0\), John and Katie both made errors. John made an error that resulted in c being incorrect while a and b were correct. Katie made an error that resulted in b being incorrect while a and c were correct. If John obtained 6 and 2 as solutions and Katie obtained -7 and -1 as solutions, and neither of them made any additional errors, what were the correct solutions to the original quadratic equation?

A. -8 and 7
B. -8 and 12
C. -6 and -2
D. 1 and 6
E. 1 and 7

is there a reason why we don't consider the option that in the quadratic equation, ax^2 + bx + c = 0 'a' could be greater than 1?
Bunuel
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John’s equation: x^2 - 8x + 12 = 0
Katie’s equation: x ^2 + 8x + 7 = 0

We’re told neither of them made a mistake in the value of a, and in both equations a = 1. So the original equation must also have a = 1.

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While rewriting a quadratic equation into the form ax^2 + bx + c = 0
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05 Jan 2026, 08:39
John
since a and b are correct, sum of roots which is b/a is also correct that means 6+2 is the true sum of roots,

Katie
since a and c are correct, the product of roots which is -c/a is correct that means, -7*(-1) is the true product of roots.

The only option which is directly visible where the sum of roots is 6+2=8 and product of roots is -7*(-1)=7 is option E.

Bunuel
While rewriting a quadratic equation into the form \(ax^2 + bx + c = 0\), John and Katie both made errors. John made an error that resulted in c being incorrect while a and b were correct. Katie made an error that resulted in b being incorrect while a and c were correct. If John obtained 6 and 2 as solutions and Katie obtained -7 and -1 as solutions, and neither of them made any additional errors, what were the correct solutions to the original quadratic equation?

A. -8 and 7
B. -8 and 12
C. -6 and -2
D. 1 and 6
E. 1 and 7

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