If the graph of y = x^2 + ax + b passes through the points : GMAT Data Sufficiency (DS) - Page 2
Check GMAT Club App Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 04 Dec 2016, 17:56

### 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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:

### Hide Tags

Intern
Joined: 28 Jan 2013
Posts: 34
Followers: 0

Kudos [?]: 20 [0], given: 20

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

### Show Tags

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)

hence n-m = 2
Intern
Joined: 18 Sep 2013
Posts: 27
Followers: 0

Kudos [?]: 0 [0], given: 1

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

### Show Tags

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?
Senior Manager
Joined: 15 Aug 2013
Posts: 328
Followers: 0

Kudos [?]: 52 [0], given: 23

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

### Show Tags

18 May 2014, 11:19
Bunuel wrote:
gabrieldoria wrote:

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?

Thanks!
Math Expert
Joined: 02 Sep 2009
Posts: 35843
Followers: 6837

Kudos [?]: 89859 [0], given: 10381

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

### Show Tags

19 May 2014, 02:12
russ9 wrote:
Bunuel wrote:
gabrieldoria wrote:

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?

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: 40
Concentration: General Management, Strategy
Followers: 1

Kudos [?]: 58 [0], given: 1

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

### Show Tags

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.

Press kudos if you wish to appreciate
Manager
Joined: 04 Jan 2014
Posts: 129
Followers: 1

Kudos [?]: 9 [0], given: 24

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

### Show Tags

17 Jun 2014, 20:53
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
Joined: 02 Sep 2009
Posts: 35843
Followers: 6837

Kudos [?]: 89859 [1] , given: 10381

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

### Show Tags

18 Jun 2014, 00:32
1
KUDOS
Expert's post
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: 14
Followers: 0

Kudos [?]: 2 [0], given: 31

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

### Show Tags

20 Oct 2014, 05:18
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
Manager
Joined: 13 Dec 2013
Posts: 58
GPA: 2.71
Followers: 0

Kudos [?]: 4 [0], given: 21

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

### Show Tags

05 Dec 2014, 09:24
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: 152
GMAT 1: 630 Q48 V29
Followers: 0

Kudos [?]: 42 [0], given: 373

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

### Show Tags

08 Dec 2014, 02:17
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

[Reveal] Spoiler:
A

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
Joined: 02 Sep 2009
Posts: 35843
Followers: 6837

Kudos [?]: 89859 [0], given: 10381

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

### Show Tags

08 Dec 2014, 03:33
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

[Reveal] Spoiler:
A

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 $$x=\sqrt{9}=3$$.
_________________
GMAT Club Legend
Joined: 09 Sep 2013
Posts: 12867
Followers: 559

Kudos [?]: 157 [0], given: 0

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

### Show Tags

09 Jan 2016, 13:43
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.
_________________
Manager
Joined: 15 Jun 2016
Posts: 50
Location: India
Concentration: Technology, Strategy
GMAT 1: 730 Q50 V39
Followers: 0

Kudos [?]: 8 [0], given: 42

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

### Show Tags

09 Jul 2016, 22:36
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 ...
Math Expert
Joined: 02 Sep 2009
Posts: 35843
Followers: 6837

Kudos [?]: 89859 [1] , given: 10381

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

### Show Tags

09 Jul 2016, 22:43
1
KUDOS
Expert's post
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.
_________________
Manager
Joined: 15 Jun 2016
Posts: 50
Location: India
Concentration: Technology, Strategy
GMAT 1: 730 Q50 V39
Followers: 0

Kudos [?]: 8 [0], given: 42

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

### Show Tags

09 Jul 2016, 22:50
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
Joined: 02 Sep 2009
Posts: 35843
Followers: 6837

Kudos [?]: 89859 [0], given: 10381

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

### Show Tags

09 Jul 2016, 22:57
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"?
_________________
Manager
Joined: 15 Jun 2016
Posts: 50
Location: India
Concentration: Technology, Strategy
GMAT 1: 730 Q50 V39
Followers: 0

Kudos [?]: 8 [0], given: 42

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

### Show Tags

09 Jul 2016, 23:50
Yes.. Other than equations where such type of questions are asked (only if you have the link readily).. thanks
If the graph of y = x^2 + ax + b passes through the points   [#permalink] 09 Jul 2016, 23:50

Go to page   Previous    1   2   [ 37 posts ]

Similar topics Replies Last post
Similar
Topics:
In the graph above, the equation of line m is y = ax + b. What is the 2 15 Nov 2016, 00:37
14 Does the line y = ax + b pass through the point (2, 5)? (1) When it 6 14 Sep 2015, 22:07
3 In how many points does the curve y=x^2-qx+p pass through x-axis? 1 15 Jul 2015, 17:40
16 In the xy-coordinate plane, line L passes through the points (b, a) an 4 01 May 2015, 01:16
1 Is x=y ? a.x^2 - y =0 b.x-1>0 1 15 Jun 2011, 10:28
Display posts from previous: Sort by