GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 23 Mar 2019, 03:46 ### GMAT Club Daily Prep

#### 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

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. ### Request Expert Reply # If the graph of y = x^2 + ax + b passes through the points

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Intern  Joined: 28 Jan 2013
Posts: 29
Re: If the graph of y = x^2 + ax + b passes through the points  [#permalink]

### Show Tags

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
Intern  Joined: 18 Sep 2013
Posts: 22
Re: If the graph of y = x^2 + ax + b passes through the points  [#permalink]

### Show Tags

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?
Manager  Joined: 15 Aug 2013
Posts: 243
Re: If the graph of y = x^2 + ax + b passes through the points  [#permalink]

### Show Tags

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 $$\sqrt{x}$$, 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?

Thanks!
Math Expert V
Joined: 02 Sep 2009
Posts: 53795
Re: If the graph of y = x^2 + ax + b passes through the points  [#permalink]

### Show Tags

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 $$\sqrt{x}$$, 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?

Thanks!

Yes, this is true for all GMAT questions.

Questions involving Viete's theorem to practice:
in-the-equation-x-2-bx-12-0-x-is-a-variable-and-b-is-a-109771.html
if-x-2-3-is-one-factor-of-the-equation-x-2-4-3-x-160524.html
if-x-2-12x-k-0-is-x-155465.html
in-the-equation-ax-2-bx-c-0-a-b-and-c-are-constants-148766.html
new-algebra-set-149349-80.html#p1200987
if-q-is-one-root-of-the-equation-x-2-18x-11c-0-where-141199.html
if-f-x-5x-2-and-g-x-x-2-12x-85-what-is-the-sum-of-all-85989.html
if-4-is-one-solution-of-the-equation-x2-3x-k-10-where-139119.html
if-r-and-s-are-the-roots-of-the-equation-x-2-bx-c-141018.html

Hope this helps.
_________________
Intern  Joined: 13 May 2014
Posts: 33
Concentration: General Management, Strategy
Re: If the graph of y = x^2 + ax + b passes through the points  [#permalink]

### Show Tags

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.

Press kudos if you wish to appreciate
Manager  Joined: 04 Jan 2014
Posts: 102
Re: If the graph of y = x^2 + ax + b passes through the points  [#permalink]

### Show Tags

mbaiseasy wrote:
Given $$F(x) = Ax^2 + Bx + C$$
We could get the difference of roots with a formula:
$$x1-x2=\sqrt{{\frac{b^2-4ac}{a^2}}}$$ with $$x1>x2$$

Should the denominator be $$a^2$$ or $$2a$$?
Math Expert V
Joined: 02 Sep 2009
Posts: 53795
Re: If the graph of y = x^2 + ax + b passes through the points  [#permalink]

### Show Tags

1
pretzel wrote:
mbaiseasy wrote:
Given $$F(x) = Ax^2 + Bx + C$$
We could get the difference of roots with a formula:
$$x1-x2=\sqrt{{\frac{b^2-4ac}{a^2}}}$$ with $$x1>x2$$

Should the denominator be $$a^2$$ or $$2a$$?

It's correct as it is:

$$x_1=\frac{-b+\sqrt{b^2-4ac}}{2a}$$;

$$x_2=\frac{-b-\sqrt{b^2-4ac}}{2a}$$;

$$x_1-x_2=\frac{-b+\sqrt{b^2-4ac}}{2a}-\frac{-b-\sqrt{b^2-4ac}}{2a}=\frac{\sqrt{b^2-4ac}}{a}=\frac{\sqrt{b^2-4ac}}{\sqrt{a^2}}$$.

Hope it's clear.
_________________
Intern  Joined: 31 May 2013
Posts: 13
Re: If the graph of y = x^2 + ax + b passes through the points  [#permalink]

### Show Tags

From 1, obtain 2 quadratic equations for $$y=x^2 + ax +b$$ by substituting (m,0) and (n,0)

$$m^2 + am + b$$ = 0
$$n^2 + am + b$$ = 0

which gives

n = (-a+$$\sqrt{a^2-4b}$$)/2a ; (-b-$$\sqrt{a^2-4b}$$)/2a
m = (-a+$$\sqrt{a^2-4b}$$)/2a ; (-b-$$\sqrt{a^2-4b}$$)/2a

since n - m > 0,

n-m = $$\sqrt{a^2-4b}$$

substitute from (1) $$a^2-4b$$ = 4

n-m = $$\sqrt{4}$$
n-m =2
Sufficient
Intern  Joined: 13 Dec 2013
Posts: 38
GPA: 2.71
Re: If the graph of y = x^2 + ax + b passes through the points  [#permalink]

### Show Tags

mbaiseasy wrote:
Given $$F(x) = Ax^2 + Bx + C$$
We could get the difference of roots with a formula:
$$x1-x2=\sqrt{{\frac{b^2-4ac}{a^2}}}$$ with $$x1>x2$$

Solution:
Apply that formula to the problem:
$$y=x^2+ax+b$$ that passes (m,0) and (n,0) with n > m

$$n-m = \sqrt{\frac{a^2-4(1)(b)}{(1)^2}}=\sqrt{a^2-4b}$$

Statement (1) gives us $$\sqrt{a^2-4b}=\sqrt{4}=2$$ SUFFICIENT.
Statement (2) gives us b = 0. Thus, INSUFFICIENT.

Formula for determining the SUM, PRODUCT and DIFFERENCE of roots of $$F(x) = Ax^2 + Bx + C$$:
http://burnoutorbreathe.blogspot.com/2012/12/sum-and-product-of-roots-of-fx.html
http://burnoutorbreathe.blogspot.com/2012/12/algebra-difference-of-roots-of-fx.html

Now why didnt i read this before I read the question lol
Manager  Joined: 31 Jul 2014
Posts: 127
GMAT 1: 630 Q48 V29 Re: If the graph of y = x^2 + ax + b passes through the points  [#permalink]

### Show Tags

EvaJager wrote:
aeros232 wrote:
If the graph of y = x^2 + ax + b passes through the points (m, 0) and (n, 0), where m < n, what is the value of n – m ?

(1) 4b = a^2 – 4

(2) b = 0

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 ?
Math Expert V
Joined: 02 Sep 2009
Posts: 53795
Re: If the graph of y = x^2 + ax + b passes through the points  [#permalink]

### Show Tags

EvaJager wrote:
aeros232 wrote:
If the graph of y = x^2 + ax + b passes through the points (m, 0) and (n, 0), where m < n, what is the value of n – m ?

(1) 4b = a^2 – 4

(2) b = 0

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;
$$\sqrt{16} = 2$$, 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 $$x=\sqrt{9}=3$$.
_________________
Intern  Joined: 15 Jun 2016
Posts: 47
Location: India
Concentration: Technology, Strategy
GMAT 1: 730 Q50 V39 Re: If the graph of y = x^2 + ax + b passes through the points  [#permalink]

### Show Tags

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 ...
Math Expert V
Joined: 02 Sep 2009
Posts: 53795
Re: If the graph of y = x^2 + ax + b passes through the points  [#permalink]

### Show Tags

1
target760gmat wrote:
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.
_________________
Intern  Joined: 15 Jun 2016
Posts: 47
Location: India
Concentration: Technology, Strategy
GMAT 1: 730 Q50 V39 Re: If the graph of y = x^2 + ax + b passes through the points  [#permalink]

### Show Tags

thanks Bunuel for clarifying my doubt... Are there any other topics other than equations in which such questions could be asked? Or if you have the link it would be helpful.. thanks
Math Expert V
Joined: 02 Sep 2009
Posts: 53795
Re: If the graph of y = x^2 + ax + b passes through the points  [#permalink]

### Show Tags

target760gmat wrote:
thanks Bunuel for clarifying my doubt... Are there any other topics other than equations in which such questions could be asked? Or if you have the link it would be helpful.. thanks

What do you mean by "such questions"? Or by "other than equations"?
_________________
Intern  Joined: 15 Jun 2016
Posts: 47
Location: India
Concentration: Technology, Strategy
GMAT 1: 730 Q50 V39 If the graph of y = x^2 + ax + b passes through the points  [#permalink]

### Show Tags

Yes.. Other than equations where such type of questions are asked (only if you have the link readily).. thanks
Intern  B
Joined: 09 May 2016
Posts: 43
Re: If the graph of y = x^2 + ax + b passes through the points  [#permalink]

### Show Tags

EvaJager wrote:
aeros232 wrote:
If the graph of y = x^2 + ax + b passes through the points (m, 0) and (n, 0), where m < n, what is the value of n – m ?

(1) 4b = a^2 – 4

(2) b = 0

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 $$a$$.

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
Math Expert V
Joined: 02 Sep 2009
Posts: 53795
Re: If the graph of y = x^2 + ax + b passes through the points  [#permalink]

### Show Tags

KARISHMA315 wrote:
EvaJager wrote:
aeros232 wrote:
If the graph of y = x^2 + ax + b passes through the points (m, 0) and (n, 0), where m < n, what is the value of n – m ?

(1) 4b = a^2 – 4

(2) b = 0

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 $$a$$.

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

You have to go through the basics before attempting questions. Check here: http://gmatclub.com/forum/coordinate-ge ... 87652.html
_________________
Manager  B
Joined: 13 Apr 2017
Posts: 55
Re: If the graph of y = x^2 + ax + b passes through the points  [#permalink]

### Show Tags

the problem is interesting, but it takes me more than 2 min to find the correct answer
Intern  B
Joined: 04 Apr 2018
Posts: 21
If the graph of y = x^2 + ax + b passes through the points  [#permalink]

### Show Tags

MGMAT official solution.
could someone please explain their explanation for the second statement? (in red below)
Bunuel

"Before looking at the statements, make sure you understand what the question is asking. In particular, what is the connection between a, b, m, and n?

The function y = x^2 + ax + b is quadratic, so it can’t intersect the x-axis in more than two places. Since (m, 0) and (n, 0) are distinct x-intercepts, they must be the only two x-intercepts of the parabola. Therefore, the equation of the parabola can be written as y = (x – m)(x – n), which can be expanded into y = x^2 – mx – nx + mn, or y = x^2 + (­–m – n)x + mn. Since this equation must be equal to the equation y = x^2 + ax + b, it follows that a = –m – n and b = mn.

(1) SUFFICIENT: We have equations that relate a and b to m and n. Replace a with (−m – n) and b with (mn):

4(mn) = (–m – n)2 – 4
4mn = m2 + 2mn + n2 – 4
4 = m2 – 2mn + n2
4 = (m – n)2
m – n = 2 or –2

Since the problem specifies that m < n, the first of these is impossible. Therefore, m – n = –2, or n – m = 2.

We can also deal with this statement by using smart numbers.

Solve the statement by dividing by 4, giving b = a^2/4 – 1. Then substitute values for a, solve for b, and then find m and n by solving the resulting quadratic.

· If a = 2, then b = 4/4 – 1 = 0. Therefore, the equation is y = x^2 + 2x, which factors to y = x(x + 2). The x-intercepts of this function are m = –2 and n = 0, so n – m = 2.

· If a = 4, then b = 16/4 – 1 = 3. Therefore, the equation is y = x^2 + 4x + 3, which factors to y = (x + 1)(x + 3). The x-intercepts of this function are m = –3 and n = –1, so n – m = 2.

· If a = 6, then b = 36/4 – 1 = 8. Therefore, the equation is y = x^2 + 6x + 8, which factors to y = (x + 2)(x + 4). The x-intercepts of this function are m = –4 and n = –2, so n – m = 2.

Try more cases if necessary (you may want to try a = 0, or a = negative); you’ll get n – m = 2 every time.

(2) INSUFFICIENT: If b = 0, then the equation is y = x2 + ax, where a is an unspecified constant. This equation factors to y = x(x + a) and so has x-intercepts –a and 0. Therefore, n – m = 0 – (–a) = a. Since a is unspecified, there are many possible values.

We can also deal with this statement by using smart numbers.

We know b = 0; pick different values for a.

· If a = 1, then the equation is y = x^2 + x, which factors to y = x(x + 1). The x-intercepts of this equation are –1 and 0, so n – m = 0 – (–1) = 1.

· If a = 2, then the equation is y = x^2 + 2x, which factors to y = x(x + 2). The x-intercepts of this equation are –2 and 0, so n – m = 0 – (–2) = 2.

We have found two different values for n – m, so this statement is insufficient.

The correct answer is A. If the graph of y = x^2 + ax + b passes through the points   [#permalink] 01 Dec 2018, 03:43

Go to page   Previous    1   2   3    Next  [ 44 posts ]

Display posts from previous: Sort by

# If the graph of y = x^2 + ax + b passes through the points

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.  