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Re: Inequalities, how to crack solve this [#permalink]

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23 Aug 2011, 23:48

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]

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24 Aug 2011, 00:06

\(x^2+ax+b\) => 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)

Re: Inequalities, how to crack solve this [#permalink]

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24 Aug 2011, 01:47

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 = \(2\sqrt{b}\)

Re: Inequalities, how to crack solve this [#permalink]

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24 Aug 2011, 12:31

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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