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Algebra> If a and b are positive numbers , prove t...

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If a and b are positive numbers , prove that the equation 1/x + (1/x+a) +(1/x+b) =0 has 2 real roots ,one b/w a/3 and 2a/3 and other b/w – 2b/3 and – b/3

muktesh singh , 10 Years ago

Grade 11

1 Answers

Jitender Singh

Last Activity: 10 Years ago

Ans:
Hello Student,
Please find the solution of your question below,
$\frac{1}{x} + \frac{1}{x + a} + \frac{1}{x + b} = 0$
$\frac{(x+a)(x+b)+x(x+b)+x(x+a)}{x.(x+a)(x+b)} = 0$
$3x^{2}+2bx+2ax+ab = 0$
$f(x) = 3x^{2}+2x(a+b)+ab = 0$
$\Delta = b^{2}-4ac$
$\Delta = 4(a+b)^{2}-4.3.ab$
$\Delta = 4(a^{2}+b^{2}+2ab)-12ab$
$\Delta = 2a^{2}+2b^{2}+2a^{2}+2b^{2}-4ab$
$\Delta = 2a^{2}+2b^{2}+2(a-b)^{2}$
$\Delta \geq 0$
-> So equation f(x) has two real roots.
$f(x) = 3x^{2}+2x(a+b)+ab$
You can easily found the value from here.

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