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

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

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

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

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New post 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?

Thanks!
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New post 19 May 2014, 02:12
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?

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
john-and-jane-started-solving-a-quadratic-equation-john-mad-106597.html
if-r-and-s-are-the-roots-of-the-equation-x-2-bx-c-141018.html

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

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New post 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.

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New post 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 \(2a\)?
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Re: If the graph of y = x^2 + ax + b passes through the points  [#permalink]

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New post 18 Jun 2014, 00:32
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.
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Re: If the graph of y = x^2 + ax + b passes through the points  [#permalink]

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

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

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New post 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



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 ?
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New post 08 Dec 2014, 03:33
anupamadw 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 \(n-m=\sqrt{4}=2.\)
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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Re: If the graph of y = x^2 + ax + b passes through the points  [#permalink]

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New post 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 ...
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New post 09 Jul 2016, 22:43
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.
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Re: If the graph of y = x^2 + ax + b passes through the points  [#permalink]

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New post 09 Jul 2016, 22:50
thanks Bunuel for clarifying my doubt... :-D

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
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New post 09 Jul 2016, 22:57
target760gmat wrote:
thanks Bunuel for clarifying my doubt... :-D

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

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

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New post 26 Mar 2017, 02:14
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 \(a\).

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
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New post 26 Mar 2017, 04:08
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 \(a\).

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


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

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New post 07 May 2017, 23:40
the problem is interesting, but it takes me more than 2 min to find the correct answer
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If the graph of y = x^2 + ax + b passes through the points  [#permalink]

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New post 01 Dec 2018, 02:43
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.
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