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Quadratic Equation Problem_2

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Quadratic Equation Problem_2

by [email protected] » Tue Jan 01, 2013 11:22 am
If the roots of the equation px^2+rx+r=0 are in the ratio a:b, then the value of \sqrt{b/a}+\sqrt{a/b} is

A. \sqrt{1/r}



B. -\sqrt{r/p}



C. \sqrt{l/p}



D. \sqrt{1/p}



E. \sqrt{3/p}


The problem has been issued in Jumbo Test; I am not sure of the answer though.Would much appreciate an expert advise for this problem.

Thanks
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by Anurag@Gurome » Wed Jan 02, 2013 12:21 am
[email protected] wrote:If the roots of the equation px² + rx + r = 0 are in the ratio a:b, then the value of √(b/a)+√(a/b) is
If m and n are the roots of the quadratic equation ax² + bx + c = 0, then
  • (m + n) = -b/a
    mn = c/a
√(b/a) + √(a/b) = (a + b)/√(ab)

Let us assume the roots of the quadratic equation px² + rx + r = 0 are ax and bx.
Hence, (ax + bx) = -r/p and abx² = r/p
Hence, (a + b) = -r/(px) and √(ab) = √(r/p)/x

Hence, √(b/a) + √(a/b) = (a + b)/√(ab) = [-r/(px)]/[√(r/p)/x] = (-r/p)/√(r/p) = -√(r/p)

The correct answer is B.
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by srinivas2357 » Thu Jan 03, 2013 4:40 am
If the roots of the equation px^2+rx+r=0 are in the ratio a:b, then the value of \sqrt{b/a}+\sqrt{a/b} is
A. \sqrt{1/r}
B. -\sqrt{r/p}
C. \sqrt{l/p}
D. \sqrt{1/p}
E. \sqrt{3/p}
let the roots of the equation are ak and bk then ak+bk = -r/p and ak*bk = r/p implies a+b=-r/pk and ab=r/pk^2 sqrt(ab)=sqrt(r/p)*1/k
Now sqrt{b/a}+ sqrt{a/b}= (a+b)/sqrt(ab)= (-r/pk)/sqrt(r/p)*1/k = -sqrt(r/p).
Hence the correct option is B.
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