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Re: If the graph of y = x^2 + ax + b passes through the points [#permalink]
21 Oct 2013, 01:21

Sam1 wrote:

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

n and m are the roots of the equation, so

(n+m)=-a and nm=b and we need to find n-m. so just solve the equation to get (n-m) term.

(n-m)^2= (n+m)^2 - 4mn

put above values

(n-m)^2= (n+m)^2 - 4mn = a^2-4b = 4 (as per the given equation, i)

Re: If the graph of y = x^2 + ax + b passes through the points [#permalink]
21 Oct 2013, 04:45

Chiranjeevee wrote:

Sam1 wrote:

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

n and m are the roots of the equation, so

(n+m)=-a and nm=b and we need to find n-m. so just solve the equation to get (n-m) term.

(n-m)^2= (n+m)^2 - 4mn

put above values

(n-m)^2= (n+m)^2 - 4mn = a^2-4b = 4 (as per the given equation, i)

hence n-m = 2

Hi thank you for your reply. While I understand the part with the sum of roots. I am still a little confused of how are we ignoring the b term. Are we equating (n-m)^2 to a^2?

Re: If the graph of y = x^2 + ax + b passes through the points [#permalink]
18 May 2014, 11:19

Bunuel wrote:

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.

Hope it helps.

Hi Bunuel,

This is news to me. Is this valid on ALL DS and PS problems or just one subset? Does this hold true for inequalities as well?

Also, can you suggest similar quadratics that use vieta's theorem and require us to relate roots to quadratics?

Re: If the graph of y = x^2 + ax + b passes through the points [#permalink]
19 May 2014, 02:12

Expert's post

russ9 wrote:

Bunuel wrote:

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.

Hope it helps.

Hi Bunuel,

This is news to me. Is this valid on ALL DS and PS problems or just one subset? Does this hold true for inequalities as well?

Also, can you suggest similar quadratics that use vieta's theorem and require us to relate roots to quadratics?

Re: If the graph of y = x^2 + ax + b passes through the points [#permalink]
19 May 2014, 02:14

Sam1 wrote:

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

Hi Sam1, Basically we need to find the value of n-m, the difference between the roots. What you have done is, showed the sum of the roots , n+m = - a ---(1) which you have deduced right. NOW, multiplying the above equation by n yields: n^2+m*n = -a*n or n^2 +m*n + a*n =0 or m*n - b = 0 ---- (as n^2+a*n+b =0 ,n being one of the roots) or m*n = b ----------(2)

Now, (n-m)^2 = n^2 -2*m*n + m^2 = (n+m)^2 - 4*m*n

using eqn(1) and eqn(2), we have (n-m)^2 = a^2 - 4*b or (n-m) = sq.root(a^2 - 4*b)

Now from answer choices,using option 1,we can find n-m = 2 however,using option 2, we can not find out n-m. Hence, A. Hope this helps.

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 \(n-m=\sqrt{4}=2.\) here we dont have to consider +2 and -2 ?

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 \(n-m=\sqrt{4}=2.\) here we dont have to consider +2 and -2 ?

Because \(\sqrt{4}=2\) ONLY, not +/-2.

When the GMAT provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the positive root. That is:

\(\sqrt{9} = 3\), NOT +3 or -3; \(\sqrt[4]{16} = 2\), NOT +2 or -2;

Notice that in contrast, the equation \(x^2 = 9\) has TWO solutions, +3 and -3. Because \(x^2 = 9\) means that \(x =-\sqrt{9}=-3\) or \(x=\sqrt{9}=3\). _________________

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