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

A short summary about the quadratic equation \(x^2+ax+b=0\). Using a little bit of algebraic manipulation, it can be written as:
\(x^2+2\cdot{\frac{a}{2}}x+\frac{a^2}{4}+b-\frac{a^2}{4}=0\) or \((x+\frac{a}{2})^2=\frac{a^2}{4}-b\).
Now, it is obvious that this equation will have real roots if and only if \(\frac{a^2}{4}-b=\frac{a^2-4b}{4}\geq{0}\) or \(a^2-4b\geq{0}\).

Taking the square root from both sides, we obtain \(|x+\frac{a}{2}|=|\frac{\sqrt{a^2-4b}}{2}|\), from which we get \(x_1=-\frac{a}{2}-\frac{\sqrt{a^2-4b}}{2}\) and \(x_2=-\frac{a}{2}+\frac{\sqrt{a^2-4b}}{2}\).

From here, we can deduce the sum of the two roots being \(x_1+x_2=-a\) and the product \(x_1x_2=b\). These two are the most frequently used formulas for the given quadratic equation. The sum is easy to get, as the two square roots cancel out. For the product, we use the formula \((m+n)(m-n)=m^2-n^2\):

\((-\frac{a}{2}+\frac{\sqrt{a^2-4b}}{2})(-\frac{a}{2}-\frac{\sqrt{a^2-4b}}{2})=(\frac{a}{2})^2-(\frac{\sqrt{a^2-4b}}{2})^2=\frac{a^2}{4}-\frac{a^2-4b}{4}=b\).

In addition, the difference of the roots can be easily deduced and it is \(\sqrt{a^2-4b}\).

On the GMAT, we seldom use the formula which gives the roots, as it is usually much easier to factorize the quadratic expression and immediately find the roots.
Try out for the equation \(x^2-4x+3=0\). Since \(x^2-4x+3=(x-3)(x-1)\), we can immediately see that the two roots are \(1\) and \(3\).
Check out the above formulas and see that you obtain the same roots, product, sum and difference.

The formula for the roots of the quadratic equation appears in the OG in the Math Review part for the general form \(ax^2+bx+c=0\).
In the above case we have \(a=1, \,\,b=a \,\,and \,\,c=b\).

Hello, Can you please explain the highlighted part ?

Basically, if you take given equation x^2+ax+b and compare with generice equation Ax^2 + Bx +C
you get A= 1, B=a, C=b
Hence some of roots =-B/A = -a
and product of roots = C/A = b

=> m+n = -a
and mn = b

What souvik has done is - he took same variables (a,b,c) in generic equation and given equation, that is incorrect and therefore caused confusion.
Hope it is clear now.

Basically, if you take given equation x^2+ax+b and compare with generice equation Ax^2 + Bx +C
you get A= 1, B=a, C=b
Hence some of roots =-B/A = -a
and product of roots = C/A = b

=> m+n = -a
and mn = b

What souvik has done is - he took same variables (a,b,c) in generic equation and given equation, that is incorrect and therefore caused confusion.
Hope it is clear now.


Kudos for you.. !!! cheers

This is a good one!!
MGMAT I think!
anyway
from the question we can understand the for \(x=m\)and for\(x=n\),\(y=0\)
So it is safe to conclude and m and n and two roots of the equation
\(x^2+ax+b\)
we know that for the equation \(ax^2+bx+c=0,\)sum of the roots is \(-b/a\) and product of the roots is\(c/a\)
So in this case,
\(m+n=-a\\
mn=b\)Statement 1
So, \((n-m)^2=(m+n)^2-4mn\\
=a^2-4b\\
=4\)
So \((n-m)^2=4\)
So \((n-m)=2 (since n>m)\)
Sufficient.

Statement 2
We only know that \(b=0\) so that \(mn=0\)
That does not help us in knowing\((n-m)\)
Insufficient
Hence A[/quote]


Hello, Can you please explain the highlighted part ?[/quote]

Basically, if you take given equation x^2+ax+b and compare with generice equation Ax^2 + Bx +C
you get A= 1, B=a, C=b
Hence some of roots =-B/A = -a
and product of roots = C/A = b

=> m+n = -a
and mn = b

What souvik has done is - he took same variables (a,b,c) in generic equation and given equation, that is incorrect and therefore caused confusion.
Hope it is clear now.


Kudos for you.. !!! cheers [/quote]


Souvik didn't do anything wrong, there is missing a comma, there was no separation between m+n=-a and mn=b.

No one is blaming Souvik
However

excerpts:
roots are m and n
sum of the roots is -b/a and product of the roots is c/a
and then m+n = -a and mn =b

It is confusing and confusion is caused by same variable a,b,c used in both. Separation by comma is not as important as confusion created with the above mentioned reason.

That's why is important to remember the content and not the exact list of variables.
The sum of the two roots is given by the opposite of the ratio between the coefficient of the linear term and that of the coefficient of the quadratic term.
The product of the two roots is given by the ratio between the the coefficient of the free term and that of the coefficient of the quadratic term.

Usually, on the GMAT, the coefficient of the quadratic term is 1, so there are no fractions involved.

Not sure what is the issue here? did I say anything else? 154238 had a problem in understanding bcaz of confusing term and I explained it.. what is the issue?

I am not geting this question..

Bunuel Or karishma..plz experts...can u explain it

sanjoo, the question is hard enough but has an elegant solution:)

I like souvik101990's solution. In this case the only thing you need to know is Vieta's formulas, m+n = -a (first); m*n = b (second).

As we need to find 'n-m', first of all we should transform 'n-m'. Let's square n-m, \((n-m)^2 = n^2 - 2mn + m^2\). Clear so far?
Then we add 2mn and subtract 2mn. As (2mn - 2mn) = 0, we did not misrepresent our equation:
\((n-m)^2 = n^2 - 2mn + m^2 +2mn - 2mn = (n^2 + 2mn + m^2) -2mn - 2mn = (n+m)^2 - 4mn\).

Given (first), \((-a)^2 = (m+n)^2\) and (second) \(- 4 * b = -4mn\), so \((n-m)^2 = a^2 - 4b\), given statement (1), \((n-m)^2 = a^2 - 4b = 4\). So \(n-m = 2\) (as n > m)

I hope it's clear.

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 \(F(x) = Ax^2 + Bx + C\):
https://burnoutorbreathe.blogspot.com/2012/12/sum-and-product-of-roots-of-fx.html
https://burnoutorbreathe.blogspot.com/2012/12/algebra-difference-of-roots-of-fx.html

dafuq just happened :D

Thanks evajager for having my back
Vips you are right, choosing variables is pretty important when trying to explain a problem

Where did you get the formula? Are you sure there's no easier way to do this problem?

Even if you don't know the formula, you can still proceed(Though it won't be a bad idea to know this formula now) and this method is actually the same thing.

So we have , from the first F.S , \(y = x^2+ax+b.\) Also, \(b = \frac{a^2-4}{4}\). Replacing this value, we have :

\(y = x^2+ax+\frac{a^2-4}{4} \to y = x^2+ax+\frac{a^2}{4} -1 \to y+1 = (a+\frac{x}{2})^2\)

Now, for y=0, we have two values of x,x=m and x=n. Thus, \(1=(x+\frac{a}{2})^2 \to (x+\frac{a}{2}) = \pm{1}\)

Hence,\(x=1-\frac{a}{2}\) and \(x = -1-\frac{a}{2}\), Again, as n>m,\(n = 1-\frac{a}{2}\) and\(m = -1-\frac{a}{2}\)

n-m = \(1-\frac{a}{2} +1+\frac{a}{2}\) = 2.Sufficient.

F.S 2 gives us y = x(x+a) and for y=0, we have the roots as x=0 and x=-a, and as the difference of the two roots will be a function of a,an unknown variable,this is Insufficient.

Hope this helps.

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.

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