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feruz77
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equation puzzle - 2 09 Dec 2010, 07:46
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Given the quadratic equation x^2-(A-3)x-(A-7), for what value of A will the sum of the squares of the roots be zero?
a) -2
b) 3
c) 6
d) 1
e) none of these



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equation puzzle - 2 09 Dec 2010, 20:13
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feruz77
Given the quadratic equation x^2-(A-3)x-(A-7), for what value of A will the sum of the squares of the roots be zero?
a) -2
b) 3
c) 6
d) 1
e) none of these
Show more

I am assuming the equation is \(x^2-(A-3)x-(A-7) = 0\)
Let us say the roots of the equation \(x^2-(A-3)x-(A-7) = 0\) are m and n.

We want the value of A for which \(m^2 + n^2 = 0\) holds.
We know that in a quadratic, Sum of roots = -b/a and product of roots = c/a (a , b and c are co-efficient of x^2, co-efficient of x and constant term respectively.)

So m + n = (A - 3) and mn = -(A - 7)

\(m^2 + n^2 = (m + n)^2 - 2mn = (A - 3)^2 + 2(A - 7)\)

\((A - 3)^2 + 2(A - 7) = A^2 - 4A - 5 = 0\)

A = 5 or -1

Answer (E)
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equation puzzle - 2 09 Dec 2010, 20:31
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Karishma are not to worry about the signs of M and N at the beginning of solving this problem? or do you assume if m is negative it still is m for the sake of solving for example?

Also you subtract -2(MN) from (m+n)^2 because this would actually equal m^2+n^2?

toughie.
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equation puzzle - 2 10 Dec 2010, 02:09
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VeritasPrepKarishma
feruz77
Given the quadratic equation x^2-(A-3)x-(A-7), for what value of A will the sum of the squares of the roots be zero?
a) -2
b) 3
c) 6
d) 1
e) none of these

I am assuming the equation is \(x^2-(A-3)x-(A-7) = 0\)
Let us say the roots of the equation \(x^2-(A-3)x-(A-7) = 0\) are m and n.

We want the value of A for which \(m^2 + n^2 = 0\) holds.
We know that in a quadratic, Sum of roots = -b/a and product of roots = c/a (a , b and c are co-efficient of x^2, co-efficient of x and constant term respectively.)

So m + n = (A - 3) and mn = -(A - 7)

\(m^2 + n^2 = (m + n)^2 - 2mn = (A - 3)^2 + 2(A - 7)\)

\((A - 3)^2 + 2(A - 7) = A^2 - 4A - 5 = 0\)

A = 5 or -1

Answer (E)
Show more

That's not correct. I don't think that Viete's formula has anything to do with this question. Moreover, if you substitute A=5 and A=-1 then you'll get two equations which have no real roots: \(x^2-2x+2=0\) and \(x^2+4x+8=0\)

The sum of the squares of the roots to be zero or \((x_1)^2+(x_2)^2=0\) to be true, then given quadratic equation must have double root equal to zero (the sum of two non-negative values to be equal to zero both must be zero).

Only \(x^2=0\) has double root equal to zero, but there is no value of A for which \(x^2-(A-3)x-(A-7)=0\) can be transformed to \(x^2=0\): A can not simultaneously equal to 7 and 3. If it were \(x^2-(A-3)x-(A-3)=0\) then the answer would be A=3 as in this case: \(x^2-(A-3)x-(A-3)=0\) --> \(x^2=0\).

Not a good question. Don't study.
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equation puzzle - 2 10 Dec 2010, 04:25
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Bunuel
VeritasPrepKarishma
feruz77
Given the quadratic equation x^2-(A-3)x-(A-7), for what value of A will the sum of the squares of the roots be zero?
a) -2
b) 3
c) 6
d) 1
e) none of these

I am assuming the equation is \(x^2-(A-3)x-(A-7) = 0\)
Let us say the roots of the equation \(x^2-(A-3)x-(A-7) = 0\) are m and n.

We want the value of A for which \(m^2 + n^2 = 0\) holds.
We know that in a quadratic, Sum of roots = -b/a and product of roots = c/a (a , b and c are co-efficient of x^2, co-efficient of x and constant term respectively.)

So m + n = (A - 3) and mn = -(A - 7)

\(m^2 + n^2 = (m + n)^2 - 2mn = (A - 3)^2 + 2(A - 7)\)

\((A - 3)^2 + 2(A - 7) = A^2 - 4A - 5 = 0\)

A = 5 or -1

Answer (E)

That's not correct. I don't think that Viete's formula has anything to do with this question. Moreover, if you substitute A=5 and A=-1 then you'll get two equations which have no real roots: \(x^2-2x+2=0\) and \(x^2+4x+8=0\)

The sum of the squares of the roots to be zero or \((x_1)^2+(x_2)^2=0\) to be true, then given quadratic equation must have double root equal to zero (the sum of two non-negative values to be equal to zero both must be zero).

Only \(x^2=0\) has double root equal to zero, but there is no value of A for which \(x^2-(A-3)x-(A-7)=0\) can be transformed to \(x^2=0\): A can not simultaneously equal to 7 and 3. If it were \(x^2-(A-3)x-(A-3)=0\) then the answer would be A=3 as in this case: \(x^2-(A-3)x-(A-3)=0\) --> \(x^2=0\).

Not a good question. Don't study.
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I disagree. There is nothing wrong with the solution. It is not given that the roots of the equation are real. It is a reduced quadratic and Viete's theorem is absolutely applicable. It includes imaginary roots.

When A = 5, equation become \(x^2 - 2x + 2=0\)
Roots of this equation are 1 + i and 1 - i which square up and add to give 0.
When A = -1, equation becomes \(x^2 + 4x + 8=0\)
Roots of this equation are -2 -2i and -2+2i which square up and add to give 0.

As I mentioned in poster's previous question, these are not from GMAT specific material. Hence this question does include complex roots. But that doesn't mean it cannot be solved.
The solution is absolutely valid but includes topics beyond GMAT.
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equation puzzle - 2 10 Dec 2010, 05:48
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gettinit
Karishma are not to worry about the signs of M and N at the beginning of solving this problem? or do you assume if m is negative it still is m for the sake of solving for example?

Also you subtract -2(MN) from (m+n)^2 because this would actually equal m^2+n^2?

toughie.
Show more

When you assume values to be m or n, you don't have to worry about the signs. e.g. m could take the value 5 or -5.

And yes, since (m+n)^2 = m^2+n^2 + 2mn, I substituted for m^2+n^2.

Nevertheless, as Bunuel pointed it, this question involved complex roots and you will not see anything such as this in GMAT. Since the square of roots add up to give you 0, if you consider only real values, the only possible option is that both roots are 0.

If I change the question a little and make it:

Given the quadratic equation x^2-(A+2)x+(A+3) = 0, for what value of A will the sum of the squares of the roots be 13?

and use the same method as above,
m^2 + n^2 = (m + n)^2 - 2mn =13
(A + 2)^2 - 2(A +3) = 13
When you solve this, you get A = -5 or 3

Equations can be:
x^2 + 3x - 2 = 0 or x^2 -5x + 6 = 0
The sum of the squares of their roots will definitely add up to give 13.
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equation puzzle - 2 10 Dec 2010, 05:53
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VeritasPrepKarishma
I disagree. There is nothing wrong with the solution. It is not given that the roots of the equation are real. It is a reduced quadratic and Viete's theorem is absolutely applicable. It includes imaginary roots.

When A = 5, equation become \(x^2 - 2x + 2=0\)
Roots of this equation are 1 + i and 1 - i which square up and add to give 0.
When A = -1, equation becomes \(x^2 + 4x + 8=0\)
Roots of this equation are -2 -2i and -2+2i which square up and add to give 0.

As I mentioned in poster's previous question, these are not from GMAT specific material. Hence this question does include complex roots. But that doesn't mean it cannot be solved.
The solution is absolutely valid but includes topics beyond GMAT.
Show more

The question is posted on GMAT problem solving subforum. Thus unless otherwise mentioned (and neither author nor your solution mentions imaginary numbers or anything about the scope of this question) we should assume that the question obey GMAT rules, which say that GMAT is dealing only with real numbers. So from this, GMAT point of view, the question is flawed and the solution is not correct as there exist no real A for which the squares of the real roots of the given equation add up to zero.

If we include imaginary numbers then yes, for A=5 the roots of \(x^2-2x+2=0\) will be \(x_1=1-i\) and \(x_2=1+i\) --> \((x_1)^2+(x_2)^2=1-2i+i^2+1+2i+i^2=2+2i^2\) as \(i^2=-1\) then \((x_1)^2+(x_2)^2=2+2i^2=0\). And for A=-1 the roots of \(x^2+4x+8=0\) will be \(x_1=-2-2i\) and \(x_2=-2+2i\) --> \((x_1)^2+(x_2)^2=4+8i+4i^2+4-8i+4i^2=0\).

So, I assumed that there was second typo (apart from not having "=0" in equation) and there should be something like \(x^2-(A-3)x-(A-3)=0\) and then simple solution would be without Viete's formula.
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equation puzzle - 2 10 Dec 2010, 07:20
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Hi Karishma and Bunuel

I did really like your discussion and analysis of this question. I appreciate your contribution.

On this question I did wanty to reveal possible solution ways.
Moreover, I agree with both of you regarding the type of this queston that it is actually improbable to get such a kind of question at the real GMAT. But when I first encountered this question I wondered whether it really is a question of GMAT format and if not then why in the MathQuestion Bank which I got from www.WinningPrep.com it was given. You have clarified my doubt, thanks.
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equation puzzle - 2 10 Dec 2010, 07:51
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feruz77
Hi Karishma and Bunuel

I did really like your discussion and analysis of this question. I appreciate your contribution.

On this question I did wanty to reveal possible solution ways.
Moreover, I agree with both of you regarding the type of this queston that it is actually improbable to get such a kind of question at the real GMAT. But when I first encountered this question I wondered whether it really is a question of GMAT format and if not then why in the MathQuestion Bank which I got from https://www.WinningPrep.com it was given. You have clarified my doubt, thanks.
Show more


The only reason that comes to mind for why a GMAT website posted this question is that here you don't actually need to find the complex roots. So it doesn't matter whether the roots of the equation are actually real or imaginary. Just solve for A and the quadratic you get in A has perfectly fine roots. Also, you could very well do it without using Viete's theorem of sum of roots and product of roots.
(x - m)(x - n) = x^2 -(m + n)x + mn = x^2-(A-3)x-(A-7)
You can substitute for (m + n) = (A - 3) and mn = -(A - 7) and get your answer.

Still, the given condition that sum of squares of roots add up to 0 could confuse people since only real roots are tested on GMAT. Hence, I will be very very surprised if GMAT gives this question as it is. The modified question I mentioned to gettinit in the post above is still fine.
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equation puzzle - 2 10 Dec 2010, 08:54
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feruz77
Hi Karishma and Bunuel

I did really like your discussion and analysis of this question. I appreciate your contribution.

On this question I did wanty to reveal possible solution ways.
Moreover, I agree with both of you regarding the type of this queston that it is actually improbable to get such a kind of question at the real GMAT. But when I first encountered this question I wondered whether it really is a question of GMAT format and if not then why in the MathQuestion Bank which I got from https://www.WinningPrep.com it was given. You have clarified my doubt, thanks.
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You are right, there is no way you'll see such question on actual GMAT. Remember GMAT deals only with real numbers.

Next, if on GMAT you'll see that the sum of the squares of a quadratic equation is zero then you should know that this quadratic equation is x^2=0 as it's the only for which the squares of the real roots add up to zero (so no Viete's formula or factorization are needed).

Anyway, as I mentioned in my first post: this is a bad question, so I wouldn't worry about it at all.



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