Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

Show Tags

21 Oct 2013, 02:21

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]

Show Tags

21 Oct 2013, 05:45

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]

Show Tags

18 May 2014, 12:19

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?

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]

Show Tags

19 May 2014, 03:14

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 \(x=\sqrt{9}=3\).
_________________

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

Show Tags

09 Jan 2016, 14:43

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

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

Show Tags

09 Jul 2016, 23:36

A doubt - In question all the provided information is in terms of variable x, constants a,b,c etc. If our objective is to able to find a solution - then by taking only (2) gives us the answer of 'a' right ? I do agree that we don't know the value of a, however according to question a is provided a constant and it will always be it.. Pls clarify ...

A doubt - In question all the provided information is in terms of variable x, constants a,b,c etc. If our objective is to able to find a solution - then by taking only (2) gives us the answer of 'a' right ? I do agree that we don't know the value of a, however according to question a is provided a constant and it will always be it.. Pls clarify ...

When a DS question asks about the value of some variable, then the statement(s) is sufficient ONLY if you can get the single numerical value of this variable.
_________________

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.

I have a very silly doubt but pls help. What in the question made you think that m and n are roots ,why cant they be a point (m,n) from which the graph passes

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.

I have a very silly doubt but pls help. What in the question made you think that m and n are roots ,why cant they be a point (m,n) from which the graph passes

We’ve given one of our favorite features a boost! You can now manage your profile photo, or avatar , right on WordPress.com. This avatar, powered by a service...

Sometimes it’s the extra touches that make all the difference; on your website, that’s the photos and video that give your content life. You asked for streamlined access...

A lot has been written recently about the big five technology giants (Microsoft, Google, Amazon, Apple, and Facebook) that dominate the technology sector. There are fears about the...

Post today is short and sweet for my MBA batchmates! We survived Foundations term, and tomorrow's the start of our Term 1! I'm sharing my pre-MBA notes...