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If there is exactly one root of the equation x^2 + ax + b

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If there is exactly one root of the equation x^2 + ax + b [#permalink] New post 23 Aug 2011, 22:17
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A
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E

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70% (01:38) correct 30% (01:20) wrong based on 23 sessions
If there is exactly one root of the equation x^2 + ax + b, where a and b are positive constants, what is b in terms of a?

A. a/2
B. a
C. 3a/2
D. a^2/2
E. a^2/4

OPEN DISCUSSION OF THIS QUESTION IS HERE: if-there-is-exactly-one-root-of-the-equation-x-2-ax-b-128527.html
[Reveal] Spoiler: OA

Last edited by Bunuel on 13 Jul 2013, 07:00, edited 2 times in total.
Renamed the topic and edited the question.
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Re: Inequalities, how to crack solve this [#permalink] New post 23 Aug 2011, 22:48
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i guess the source is manhattan...anywayz
i got struck with a simple technique...picking up numbers..
consider (x+1)^2=x^2+2x+1
(x+2)^2=x^2+4x+4
....similarly u can go for other numbers as well..then analyse the values of a and b by considering the given options..
ans e
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Re: Inequalities, how to crack solve this [#permalink] New post 23 Aug 2011, 23:06
x^2+ax+b => has only 1 solution
So the equation should be of the form x^2 + 2Bx + B^2=0 (i.e) (x+B)^2=0 OR x^2 - 2Bx + B^2=0 (i.e) (x-B)^2=0
Hence substitute in place of B^2 = b and in place of 2B = a
we get a=2\sqrt{b}
=> a^2=4b [Squaring on both sides]
=> b=a^2/4
(E)

Hope this helps !
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Re: Inequalities, how to crack solve this [#permalink] New post 23 Aug 2011, 23:11
how did you take a=2sqrtb, from the above equation ?
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Re: Inequalities, how to crack solve this [#permalink] New post 24 Aug 2011, 00:47
naaga wrote:
how did you take a=2sqrtb, from the above equation ?

Since the given equation is a quadratic equation, there would always be 2 roots (solution) for it. eg: (x+p)(x+q)
But the question says the equation has only 1 solution. This could be possible only when both the roots are same. i.e (x+p)(x+p).
This can be written as (x+p)^2
We know the formula for this i.e (x+p)^2 = x^2+2px+p^2
Now visualize the given equation (x^2+ax+b) with the above formula:
Coefficient of x^2 = 1
Coefficient of x = a (which is nothing but 2p)
Constant term (i.e the p^2) = b
Hence we can write:
p^2=b
=> p=\sqrt{b}
=> a=2p = 2\sqrt{b}
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Re: Inequalities, how to crack solve this [#permalink] New post 24 Aug 2011, 02:36
thank you srivats212, nice explanation.
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Re: Inequalities, how to crack solve this [#permalink] New post 24 Aug 2011, 11:31
the root must be -a/2. (Quadratic equation with only one root)
Substituting the -a/2 in the equation & equating it to 0.
You will get b= a^2 / 4
Hence, E is the answer.
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Re: Inequalities, how to crack solve this [#permalink] New post 24 Aug 2011, 20:02
one root for a quadratic equation ax^2+bx+c is possible only when b^2 = 4ac ---1

Here b = a
c= b
a = 1

substituting these values in 1, we have

a^2 = 4b => b =a^2/4

Answer is E.
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Re: Inequalities, how to crack solve this [#permalink] New post 26 Aug 2011, 05:16
Can be possible only when determinent = 0

therefore, b ^2 - 4ac= 0
b=a , a=1 and c=b. After substituting values we will get a^2 - 4 (1)(b)=0 =>a ^2 =4b =>b= a^2/4.
Re: Inequalities, how to crack solve this   [#permalink] 26 Aug 2011, 05:16
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