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We are given that \(m\) and \(n\) are the roots of the quadratic equation \(x^2+ax+b=0.\) They are given by the formula \(x_{1,2}=\frac{-a\pm\sqrt{a^2-4b}}{2}\). The larger root \(n\) is that with a plus in front of the square root, therefore \(n-m=\frac{a}{2}+\frac{\sqrt{a^2-4b}}{2}-(\frac{a}{2}-\frac{\sqrt{a^2-4b}}{2})=\sqrt{a^2-4b}\).

(1) Sufficient, because \(a^2-4b=4\), so \(n-m=\sqrt{4}=2.\) (2) The quadratic equation becomes \(x^2+ax=0\) or \(x(x+a)=0.\) Not sufficient, as one of the roots is \(0\) and the other one is \(-a\), and we have no information about \(a\).

Answer A. _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: quadratic equations: quadratic intercepts [#permalink]
08 Oct 2012, 10:23

8

This post received KUDOS

Expert's post

1

This post was BOOKMARKED

clearmountain wrote:

If the graph of y = x2 + ax + b passes through the points (m, 0) and (n, 0), where m < n, what is the value of n – m ?

(1) 4b = a2 – 4

(2) b = 0 Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient. Both statements TOGETHER are sufficient, but NEITHER one ALONE is sufficient. EACH statement ALONE is sufficient. Statements (1) and (2) TOGETHER are NOT sufficient.

Answer is A;

Also could someone tell me if this is a realistic gmat problem from the official test? it seems too technical to be so.

This is a good one!! MGMAT I think! anyway from the question we can understand the for \(x=m\)and for\(x=n\),\(y=0\) So it is safe to conclude and m and n and two roots of the equation \(x^2+ax+b\) we know that for the equation \(ax^2+bx+c=0,\)sum of the roots is \(-b/a\) and product of the roots is\(c/a\) So in this case, \(m+n=-a mn=b\) Statement 1 So, \((n-m)^2=(m+n)^2-4mn =a^2-4b =4\) So \((n-m)^2=4\) So \((n-m)=2 (since n>m)\) Sufficient.

Statement 2 We only know that \(b=0\) so that \(mn=0\) That does not help us in knowing\((n-m)\) Insufficient Hence A _________________

Re: quadratic equations: quadratic intercepts [#permalink]
15 Oct 2012, 09:56

5

This post received KUDOS

aditi1903 wrote:

clearmountain wrote:

If the graph of y = x2 + ax + b passes through the points (m, 0) and (n, 0), where m < n, what is the value of n – m ?

(1) 4b = a2 – 4

(2) b = 0 Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient. Both statements TOGETHER are sufficient, but NEITHER one ALONE is sufficient. EACH statement ALONE is sufficient. Statements (1) and (2) TOGETHER are NOT sufficient.

Answer is A;

Also could someone tell me if this is a realistic gmat problem from the official test? it seems too technical to be so.

How is it possible only by A..im unable to understand it..can anybody explain?

A short summary about the quadratic equation \(x^2+ax+b=0\). Using a little bit of algebraic manipulation, it can be written as: \(x^2+2\cdot{\frac{a}{2}}x+\frac{a^2}{4}+b-\frac{a^2}{4}=0\) or \((x+\frac{a}{2})^2=\frac{a^2}{4}-b\). Now, it is obvious that this equation will have real roots if and only if \(\frac{a^2}{4}-b=\frac{a^2-4b}{4}\geq{0}\) or \(a^2-4b\geq{0}\).

Taking the square root from both sides, we obtain \(|x+\frac{a}{2}|=|\frac{\sqrt{a^2-4b}}{2}|\), from which we get \(x_1=-\frac{a}{2}-\frac{\sqrt{a^2-4b}}{2}\) and \(x_2=-\frac{a}{2}+\frac{\sqrt{a^2-4b}}{2}\).

From here, we can deduce the sum of the two roots being \(x_1+x_2=-a\) and the product \(x_1x_2=b\). These two are the most frequently used formulas for the given quadratic equation. The sum is easy to get, as the two square roots cancel out. For the product, we use the formula \((m+n)(m-n)=m^2-n^2\):

In addition, the difference of the roots can be easily deduced and it is \(\sqrt{a^2-4b}\).

On the GMAT, we seldom use the formula which gives the roots, as it is usually much easier to factorize the quadratic expression and immediately find the roots. Try out for the equation \(x^2-4x+3=0\). Since \(x^2-4x+3=(x-3)(x-1)\), we can immediately see that the two roots are \(1\) and \(3\). Check out the above formulas and see that you obtain the same roots, product, sum and difference.

The formula for the roots of the quadratic equation appears in the OG in the Math Review part for the general form \(ax^2+bx+c=0\). In the above case we have \(a=1, \,\,b=a \,\,and \,\,c=b\). _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: quadratic equations: quadratic intercepts [#permalink]
15 Oct 2012, 22:49

3

This post received KUDOS

souvik101990 wrote:

clearmountain wrote:

If the graph of y = x2 + ax + b passes through the points (m, 0) and (n, 0), where m < n, what is the value of n – m ?

(1) 4b = a2 – 4

(2) b = 0 Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient. Both statements TOGETHER are sufficient, but NEITHER one ALONE is sufficient. EACH statement ALONE is sufficient. Statements (1) and (2) TOGETHER are NOT sufficient.

Answer is A;

Also could someone tell me if this is a realistic gmat problem from the official test? it seems too technical to be so.

This is a good one!! MGMAT I think! anyway from the question we can understand the for \(x=m\)and for\(x=n\),\(y=0\) So it is safe to conclude and m and n and two roots of the equation \(x^2+ax+b\) we know that for the equation \(ax^2+bx+c=0,\)sum of the roots is \(-b/a\) and product of the roots is\(c/a\) So in this case, \(m+n=-a mn=b\)Statement 1 So, \((n-m)^2=(m+n)^2-4mn =a^2-4b =4\) So \((n-m)^2=4\) So \((n-m)=2 (since n>m)\) Sufficient.

Statement 2 We only know that \(b=0\) so that \(mn=0\) That does not help us in knowing\((n-m)\) Insufficient Hence A

Hello, Can you please explain the highlighted part ?

Re: quadratic equations: quadratic intercepts [#permalink]
16 Oct 2012, 00:44

2

This post received KUDOS

154238 wrote:

souvik101990 wrote:

clearmountain wrote:

If the graph of y = x2 + ax + b passes through the points (m, 0) and (n, 0), where m < n, what is the value of n – m ?

(1) 4b = a2 – 4

(2) b = 0 Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient. Both statements TOGETHER are sufficient, but NEITHER one ALONE is sufficient. EACH statement ALONE is sufficient. Statements (1) and (2) TOGETHER are NOT sufficient.

Answer is A;

Also could someone tell me if this is a realistic gmat problem from the official test? it seems too technical to be so.

This is a good one!! MGMAT I think! anyway from the question we can understand the for \(x=m\)and for\(x=n\),\(y=0\) So it is safe to conclude and m and n and two roots of the equation \(x^2+ax+b\) we know that for the equation \(ax^2+bx+c=0,\)sum of the roots is \(-b/a\) and product of the roots is\(c/a\) So in this case, \(m+n=-a mn=b\)Statement 1 So, \((n-m)^2=(m+n)^2-4mn =a^2-4b =4\) So \((n-m)^2=4\) So \((n-m)=2 (since n>m)\) Sufficient.

Statement 2 We only know that \(b=0\) so that \(mn=0\) That does not help us in knowing\((n-m)\) Insufficient Hence A

Hello, Can you please explain the highlighted part ?

Basically, if you take given equation x^2+ax+b and compare with generice equation Ax^2 + Bx +C you get A= 1, B=a, C=b Hence some of roots =-B/A = -a and product of roots = C/A = b

=> m+n = -a and mn = b

What souvik has done is - he took same variables (a,b,c) in generic equation and given equation, that is incorrect and therefore caused confusion. Hope it is clear now. _________________

Re: quadratic equations: quadratic intercepts [#permalink]
16 Oct 2012, 00:48

1

This post received KUDOS

154238 wrote:

souvik101990 wrote:

clearmountain wrote:

If the graph of y = x2 + ax + b passes through the points (m, 0) and (n, 0), where m < n, what is the value of n – m ?

(1) 4b = a2 – 4

(2) b = 0 Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient. Both statements TOGETHER are sufficient, but NEITHER one ALONE is sufficient. EACH statement ALONE is sufficient. Statements (1) and (2) TOGETHER are NOT sufficient.

Answer is A;

Also could someone tell me if this is a realistic gmat problem from the official test? it seems too technical to be so.

This is a good one!! MGMAT I think! anyway from the question we can understand the for \(x=m\)and for\(x=n\),\(y=0\) So it is safe to conclude and m and n and two roots of the equation \(x^2+ax+b\) we know that for the equation \(ax^2+bx+c=0,\)sum of the roots is \(-b/a\) and product of the roots is\(c/a\) So in this case, \(m+n=-a mn=b\)Statement 1 So, \((n-m)^2=(m+n)^2-4mn =a^2-4b =4\) So \((n-m)^2=4\) So \((n-m)=2 (since n>m)\) Sufficient.

Statement 2 We only know that \(b=0\) so that \(mn=0\) That does not help us in knowing\((n-m)\) Insufficient Hence A

Hello, Can you please explain the highlighted part ?

Basically, if you take given equation x^2+ax+b and compare with generice equation Ax^2 + Bx +C you get A= 1, B=a, C=b Hence some of roots =-B/A = -a and product of roots = C/A = b

=> m+n = -a and mn = b

What souvik has done is - he took same variables (a,b,c) in generic equation and given equation, that is incorrect and therefore caused confusion. Hope it is clear now.

Re: quadratic equations: quadratic intercepts [#permalink]
16 Oct 2012, 01:31

This is a good one!! MGMAT I think! anyway from the question we can understand the for \(x=m\)and for\(x=n\),\(y=0\) So it is safe to conclude and m and n and two roots of the equation \(x^2+ax+b\) we know that for the equation \(ax^2+bx+c=0,\)sum of the roots is \(-b/a\) and product of the roots is\(c/a\) So in this case, \(m+n=-a mn=b\)Statement 1 So, \((n-m)^2=(m+n)^2-4mn =a^2-4b =4\) So \((n-m)^2=4\) So \((n-m)=2 (since n>m)\) Sufficient.

Statement 2 We only know that \(b=0\) so that \(mn=0\) That does not help us in knowing\((n-m)\) Insufficient Hence A[/quote]

Hello, Can you please explain the highlighted part ?[/quote]

Basically, if you take given equation x^2+ax+b and compare with generice equation Ax^2 + Bx +C you get A= 1, B=a, C=b Hence some of roots =-B/A = -a and product of roots = C/A = b

=> m+n = -a and mn = b

What souvik has done is - he took same variables (a,b,c) in generic equation and given equation, that is incorrect and therefore caused confusion. Hope it is clear now.

Kudos for you.. !!! cheers [/quote]

Souvik didn't do anything wrong, there is missing a comma, there was no separation between m+n=-a and mn=b. _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: quadratic equations: quadratic intercepts [#permalink]
16 Oct 2012, 01:51

EvaJager wrote:

This is a good one!! MGMAT I think! anyway from the question we can understand the for \(x=m\)and for\(x=n\),\(y=0\) So it is safe to conclude and m and n and two roots of the equation \(x^2+ax+b\) we know that for the equation \(ax^2+bx+c=0,\)sum of the roots is \(-b/a\) and product of the roots is\(c/a\) So in this case, \(m+n=-a mn=b\)Statement 1 So, \((n-m)^2=(m+n)^2-4mn =a^2-4b =4\) So \((n-m)^2=4\) So \((n-m)=2 (since n>m)\) Sufficient.

Statement 2 We only know that \(b=0\) so that \(mn=0\) That does not help us in knowing\((n-m)\) Insufficient Hence A

Hello, Can you please explain the highlighted part ?

Basically, if you take given equation x^2+ax+b and compare with generice equation Ax^2 + Bx +C you get A= 1, B=a, C=b Hence some of roots =-B/A = -a and product of roots = C/A = b

=> m+n = -a and mn = b

What souvik has done is - he took same variables (a,b,c) in generic equation and given equation, that is incorrect and therefore caused confusion. Hope it is clear now.

Kudos for you.. !!! cheers

Souvik didn't do anything wrong, there is missing a comma, there was no separation between m+n=-a and mn=b.

No one is blaming Souvik However

excerpts: roots are m and n sum of the roots is -b/a and product of the roots is c/a and then m+n = -a and mn =b

It is confusing and confusion is caused by same variable a,b,c used in both. Separation by comma is not as important as confusion created with the above mentioned reason. _________________

Re: quadratic equations: quadratic intercepts [#permalink]
16 Oct 2012, 03:37

Vips0000 wrote:

EvaJager wrote:

This is a good one!! MGMAT I think! anyway from the question we can understand the for \(x=m\)and for\(x=n\),\(y=0\) So it is safe to conclude and m and n and two roots of the equation \(x^2+ax+b\) we know that for the equation \(ax^2+bx+c=0,\)sum of the roots is \(-b/a\) and product of the roots is\(c/a\) So in this case, \(m+n=-a mn=b\)Statement 1 So, \((n-m)^2=(m+n)^2-4mn =a^2-4b =4\) So \((n-m)^2=4\) So \((n-m)=2 (since n>m)\) Sufficient.

Statement 2 We only know that \(b=0\) so that \(mn=0\) That does not help us in knowing\((n-m)\) Insufficient Hence A

Hello, Can you please explain the highlighted part ?

Basically, if you take given equation x^2+ax+b and compare with generice equation Ax^2 + Bx +C you get A= 1, B=a, C=b Hence some of roots =-B/A = -a and product of roots = C/A = b

=> m+n = -a and mn = b

What souvik has done is - he took same variables (a,b,c) in generic equation and given equation, that is incorrect and therefore caused confusion. Hope it is clear now.

Kudos for you.. !!! cheers

Souvik didn't do anything wrong, there is missing a comma, there was no separation between m+n=-a and mn=b.

No one is blaming Souvik However

excerpts: roots are m and n sum of the roots is -b/a and product of the roots is c/a and then m+n = -a and mn =b

It is confusing and confusion is caused by same variable a,b,c used in both. Separation by comma is not as important as confusion created with the above mentioned reason.

That's why is important to remember the content and not the exact list of variables. The sum of the two roots is given by the opposite of the ratio between the coefficient of the linear term and that of the coefficient of the quadratic term. The product of the two roots is given by the ratio between the the coefficient of the free term and that of the coefficient of the quadratic term.

Usually, on the GMAT, the coefficient of the quadratic term is 1, so there are no fractions involved. _________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: quadratic equations: quadratic intercepts [#permalink]
16 Oct 2012, 03:57

EvaJager wrote:

Vips0000 wrote:

EvaJager wrote:

This is a good one!! MGMAT I think! anyway from the question we can understand the for \(x=m\)and for\(x=n\),\(y=0\) So it is safe to conclude and m and n and two roots of the equation \(x^2+ax+b\) we know that for the equation \(ax^2+bx+c=0,\)sum of the roots is \(-b/a\) and product of the roots is\(c/a\) So in this case, \(m+n=-a mn=b\)Statement 1 So, \((n-m)^2=(m+n)^2-4mn =a^2-4b =4\) So \((n-m)^2=4\) So \((n-m)=2 (since n>m)\) Sufficient.

Statement 2 We only know that \(b=0\) so that \(mn=0\) That does not help us in knowing\((n-m)\) Insufficient Hence A

Hello, Can you please explain the highlighted part ?

Basically, if you take given equation x^2+ax+b and compare with generice equation Ax^2 + Bx +C you get A= 1, B=a, C=b Hence some of roots =-B/A = -a and product of roots = C/A = b

=> m+n = -a and mn = b

What souvik has done is - he took same variables (a,b,c) in generic equation and given equation, that is incorrect and therefore caused confusion. Hope it is clear now.

Kudos for you.. !!! cheers

Souvik didn't do anything wrong, there is missing a comma, there was no separation between m+n=-a and mn=b.

No one is blaming Souvik However

excerpts: roots are m and n sum of the roots is -b/a and product of the roots is c/a and then m+n = -a and mn =b

It is confusing and confusion is caused by same variable a,b,c used in both. Separation by comma is not as important as confusion created with the above mentioned reason.

That's why is important to remember the content and not the exact list of variables. The sum of the two roots is given by the opposite of the ratio between the coefficient of the linear term and that of the coefficient of the quadratic term. The product of the two roots is given by the ratio between the the coefficient of the free term and that of the coefficient of the quadratic term.

Usually, on the GMAT, the coefficient of the quadratic term is 1, so there are no fractions involved.

Not sure what is the issue here? did I say anything else? 154238 had a problem in understanding bcaz of confusing term and I explained it.. what is the issue? _________________

Re: If the graph of y = x^2 + ax + b passes through the points [#permalink]
20 Nov 2012, 01:16

1

This post received KUDOS

sanjoo wrote:

I am not geting this question.. Bunuel Or karishma..plz experts...can u explain it

sanjoo, the question is hard enough but has an elegant solution:)

I like souvik101990's solution. In this case the only thing you need to know is Vieta's formulas, m+n = -a (first); m*n = b (second).

As we need to find 'n-m', first of all we should transform 'n-m'. Let's square n-m, \((n-m)^2 = n^2 - 2mn + m^2\). Clear so far? Then we add 2mn and subtract 2mn. As (2mn - 2mn) = 0, we did not misrepresent our equation: \((n-m)^2 = n^2 - 2mn + m^2 +2mn - 2mn = (n^2 + 2mn + m^2) -2mn - 2mn = (n+m)^2 - 4mn\).

Given (first), \((-a)^2 = (m+n)^2\) and (second) \(- 4 * b = -4mn\), so \((n-m)^2 = a^2 - 4b\), given statement (1), \((n-m)^2 = a^2 - 4b = 4\). So \(n-m = 2\) (as n > m)

Re: If the graph of y = x^2 + ax + b passes through the points [#permalink]
14 Sep 2013, 10:22

1

This post received KUDOS

Expert's post

haotian87 wrote:

Where did you get the formula? Are you sure there's no easier way to do this problem?

Even if you don't know the formula, you can still proceed(Though it won't be a bad idea to know this formula now) and this method is actually the same thing.

So we have , from the first F.S , \(y = x^2+ax+b.\) Also, \(b = \frac{a^2-4}{4}\). Replacing this value, we have :

F.S 2 gives us y = x(x+a) and for y=0, we have the roots as x=0 and x=-a, and as the difference of the two roots will be a function of a,an unknown variable,this is Insufficient.

Re: If the graph of y = x^2 + ax + b passes through the points [#permalink]
20 Oct 2013, 11:36

2

This post received KUDOS

Expert's post

gabrieldoria wrote:

Sorry I´ve got one last question after reading the whole thread.

Wouldn´t option 1 be insufficient too, because the SQ ROOT of 4 be +- 2? That would give two possible values for n-m.

No, \(\sqrt{4}=2\), not +2 or -2.

When the GMAT provides the square root sign for an even root, such as \(\sqrt{x}\) or \(\sqrt[4]{x}\), then the only accepted answer is the positive root. That is, \(\sqrt{25}=5\), NOT +5 or -5.

In contrast, the equation \(x^2=25\) has TWO solutions, +5 and -5. Even roots have only non-negative value on the GMAT.

Re: If the graph of y = x^2 + ax + b passes through the points [#permalink]
20 Oct 2013, 14:54

This is how I am solving it. Will appreciate your help to tell me where I am going wrong. since m and n are the two points through which the graph passes they will both satisfy the equation. Hence 0=m^2+am+b 0=n^2+an+b 1-2 gives 0=n^2-m^2 +a (n-m) taking n-m common (n-m) (n+m+a)=0 hence (n-m)=0 or (n+m+a)=0 n=m; (n+m)=-a looking at statement a a^2-4=4b a will be something in terms of b but we will not the exact value of a. Please tell me where am I making a mistake

gmatclubot

Re: If the graph of y = x^2 + ax + b passes through the points
[#permalink]
20 Oct 2013, 14:54

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