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Bunuel
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Bunuel
If there is exactly one solution to the equation 25x^2 − bx + 64 = 0, where b > 0, what is the value of b?

A. 26
B. 40
C. 52
D. 80
E. 100

Quadratic equation takes only one X when it looks like (x-n)^2 or (x+n)^2. In our case we have (ax-n)^2 => (ax)^2-2anx+n^2 => bx = 2*5*8*x = 80x. Answer (D)
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Quote:

The expression \(b^2 - 4ac\) is the "discriminant" of the quadratic equation.

If \(b^2 - 4ac > 0\), there are two solutions

If \(b^2 - 4ac = 0\), there is one solution

If \(b^2 - 4ac < 0\), there are no solutions

Hello fskilnik !

Could you please explain to me if we can use this approach just then the questions says that the eq has one, two solutions?

Is there any source where I can learn more about this concept?

Kind regards!
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jfranciscocuencag
Quote:

The expression \(b^2 - 4ac\) is the "discriminant" of the quadratic equation.

If \(b^2 - 4ac > 0\), there are two solutions

If \(b^2 - 4ac = 0\), there is one solution

If \(b^2 - 4ac < 0\), there are no solutions

Hello fskilnik !

Could you please explain to me if we can use this approach just then the questions says that the eq has one, two solutions?

Is there any source where I can learn more about this concept?

Kind regards!

Hi jfranciscocuencag,

Thanks for your interest in GMATH´s opinion on this matter.

The discriminant (aka "delta") of a second-degree expression \(ax^2+bx+c\) (where \(a \neq 0, b, c\) are constants) is VERY connected to the number of (real) roots of the corresponding second-degree equation (\(ax^2+bx+c = 0\)). Therefore not only in GMAT´s perpective, but in math in general, this is the common reason for using this "tool".

It is not necessary to have a question asking/saying about the number of solutions *explicitly*, tough.

Example: Is \(x^2+c > 0\) (\(c\) given constant) for every (real) value of \(x\)?
(1) \(c>0\)
(2) ...

Although an "intermediate" (at least) student will NOT use the discriminant in this case, because he/she knows that \(x^2\) is never negative and when added to \(c\), c> 0, we have a "YES", one MAY used it:
delta = (0)^2-4(1)(c) < 0 when c>0, hence we know \(x^2+c=0\) has no (real) roots for c>0, hence the expression x^2+c is always negative, or always positive. It´s the second case, of course (think about x=0, for instance)...

An even more mature student knows the behavior of the function \(y=x^2+c\) for every given value of \(c\), hence the sufficiency of the statement (1) is found even faster than in the first approach!

I hope you got the point!

Experts are able to create very high-level problems that use this kind of reasoning and, of course, we deal with many of them in our own course (my own "monsters", LoL). As you may have already realized (following us), we are increasing the number of exercises on our course on a daily basis (some new ones are presented here), so that I must say (without being modest at all) that the best source of problems is at the link you find in the footer... :)

(Our marketing guy is working hard to be able to convince potential students that the GMATH method can make all the difference, but most students prefer "softer approaches"...)

Now still doing some propaganda, our videos related to second-degree expressions also present MANY other things, some of them focused in over-90-percentile ambitious students (true), but we believe we must give our students ALL tools they may need, not only the most common ones. This is not a financially-wise approach (most students are not willing to work hard, let´s face it!), but it is part of our "excellence compromise", so to speak.

On the other hand, if you google (for instance) "delta second-degree equation" you will find the basics, examples, etc. And other (good) GMAT preparation materials/companies will also explain to you (at least) enough for the easy-intermediate questions, we must agree.

I hope we could help!

Regards and success in your studies,
Fabio.
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Bunuel
If there is exactly one solution to the equation 25x^2 − bx + 64 = 0, where b > 0, what is the value of b?

A. 26
B. 40
C. 52
D. 80
E. 100

In order to have one solution, when factoring into (blahblahblah)*(blahblahblah) = 0, the first blahblahblah and second blahblahblah must be the same.

We can tell by glancing at our equation that the way to get 25 as the x^2 coefficient is to have 5x and the way to get 64 as our x^0 coefficient is to have 8. If they are both positive or both negative, the x^1 coefficient will be positive. We don't want that, so we need one of them to be negative.

\((5x^2-8)(5x^2-8) = 0\)

\(25x^2-80x+64 = 0\)

Answer choice D.
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If there is exactly one solution to the equation 25x^2 − bx + 64 = 0, where b > 0, what is the value of b?

If there is only one solution(both the roots are equal), therefore D=0
=> 25x^2 − bx + 64 = 0
=> x^2 - (b/25)x + 64/25 = 0

For D = 0,
(b/25)^2 - 4(1)(64/25)=0
=> (b/25)^2 = (256/25) = (16/5)^2

Since b is positive,
=>(b/25) = (16/5)
=> b=80

Hence D
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You can actually spot an remarkable identity here:

(a-b)^2 = a^2 -2*a*b + b^2

as you have:
25x^2 - bx + 64

you can factor as:
(5x - 8)^2
thus you can imply that the middle value, i.e. b, is going to be 2*5*8 = 80
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Whenever you see questions involving - no. of roots : first think of using this concept :
Sum of roots = -b/a
Product of roots : c/a
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