While rewriting a quadratic equation into the form ax^2 + bx + c = 0

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

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

­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!

 
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. 

­Standard quadratic business:

Thank you both so much! This helped a lot, really appreciate you both.

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?­

­
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

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.