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Two students were asked to solve the equation

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by Max@Math Revolution » Wed Jan 17, 2018 12:00 am
Two students were asked to solve the equation $$x^2+px+q=0$$ . Alice's answer was x=1 and 5.
This answer is incorrect for the original equation, but correct for the equation $$x^2+px+r=0$$ .
Bob's answer was x=-2 and -4. This answer is incorrect for the original equation, but correct for $$x^2+sx+q=0$$ .
What are the solutions of the original equation?

A. 1, 8
B. -1,-8
C. 2,4
D. -1,-5
E. 1,-5
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by DrMaths » Wed Jan 17, 2018 5:00 am
Two students were asked to solve the equation x^2+px+q=0 .
Alice: x^2+px+r=(x-1)(x-5)=x^2+px+r = x^2-6x+10, so p=-6, r=10
Bob: x^2+sx+q = (x+2)(x+4) = x^2+6x+8, so s=6, q=8.
Substituting p=-6 and q=8 into the original equation gives: x^2-6x+8 =(x-2)(x-4)=0
So x = 2,4
ANSWER=C
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Alice and Bob

by GMATGuruNY » Wed Jan 17, 2018 5:18 am
Max@Math Revolution wrote:Two students were asked to solve the equation $$x^2+px+q=0$$ . Alice's answer was x=1 and 5.
This answer is incorrect for the original equation, but correct for the equation $$x^2+px+r=0$$ .
Bob's answer was x=-2 and -4. This answer is incorrect for the original equation, but correct for $$x^2+sx+q=0$$ .
What are the solutions of the original equation?

A. 1, 8
B. -1,-8
C. 2,4
D. -1,-5
E. 1,-5
Alternate approach:

For any equation of the form x² + bx + c = 0:
Sum of the roots = -b.
Product of the roots = c.

Original equation: x² + px + q = 0
Alice's equation: x² + px + r = 0
In each case, the sum of the roots is equal to -p.
Thus, the sum of the roots for the original equation must be equal to the sum of the roots for Alice's equation:
1+5 = 6.

Original equation: x² + px + q = 0
Bob's equation: x² + sx + q = 0
In each case, the product of the roots is equal to q.
Thus, the product of the roots for the original equation must be equal to the product of the roots for Bob's equation:
(-2)(-4) = 8.

The correct answer must offer roots with a SUM OF 6 and a PRODUCT OF 8.
Only C works:
2+4 = 6.
2*4 = 8.

The correct answer is C.
Last edited by GMATGuruNY on Wed Jan 17, 2018 9:47 am, edited 1 time in total.
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by DrMaths » Wed Jan 17, 2018 9:25 am
A variation on my previous entry.

Alice: x^2+px+r=(x-1)(x-5)=x^2+px+r = x^2-6x+10, so p=-6, r=10
Bob: x^2+sx+q = (x+2)(x+4) = x^2+6x+8, so s=6, q=8.

Now test the given answers:
A. 1, 8
E. 1,-5
- when x = 1, p+q = -1 does not compute

B. -1,-8
D. -1,-5
-when x = -1, q-p=1 does not compute

C. 2,4 MUST BE CORRECT

ANSWER=C
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EDIT

by Max@Math Revolution » Fri Jan 19, 2018 12:12 am
=>

Using the factor theorem with Alice's solution yields
(x-1)(x-5) = x^2 - 6x + 5 = x^2+px+r.
So, p = -6.

Using the factor theorem with Bob's solution yields
(x+2)(x+4) = x^2 + 6x + 8 = x^2 +sx +q.
So, q = 8.

Thus, the original equation is x^2-6x+8 = (x-2)(x-4) = 0. Its roots are 2 and 4.

Therefore, the answer is C.
Answer: C
Last edited by Max@Math Revolution on Thu Jan 25, 2018 11:15 pm, edited 1 time in total.
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by Max@Math Revolution » Sun Jan 21, 2018 8:54 am
Max@Math Revolution wrote:=>

Using the factor theorem with Alice's solution yields
(x-1)(x-5) = x^2 - 6x + 5 = x^2+px+r.
So, p = -6.

Using the factor theorem with Bob's solution yields
(x+2)(x+4) = x^2 + 6x + 8 = x^2 +sx +q.
So, q = 8.

Thus, the original equation is x^2-6x+8 = (x-2)(x-4) = 0. Its roots are 2 and 4.

Therefore, the answer is C.
Answer: C
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