Question

if alpha,beta are the roots of equation 1/x+a + 1/x+b =1/c and alpha+beta =0 show that alphabeta = -(asquare + b square/2 )

Answer 1

Answer:

$\red {\alpha \beta } \green {= \frac{-(a^{2}+b^{2})}{2}}$

Step-by-step explanation:

$\frac{1}{x+a} + \frac{1}{x+b}= \frac{1}{c}$

$\implies \frac{x+b+x+a}{(x+a)(x+b)} = \frac{1}{c}$

$\implies \frac{(2x+a+b)c}{x^{2}+(a+b)x+ab} = 1$

$\implies 2cx + (a+b)c = x^{2} + (a+b)x + ab$

$\implies x^{2} + (a+b-2c)x + [ab- (a+b)c ]= 0$

/* Compare above equation with Ax^{2}+Bx+C=0, we get

$A = 1 , B = (a+b-2c) , C = ab-(a+b)c$

$It \: is \: given \:that \: \alpha \:and \:\beta \\are \: roots \: of \: Quadratic\: equation$

$i) \alpha + \beta = 0\:(given)$

$\implies \frac{-B}{A} = 0\: (given)$

$\implies \frac{-(a+b-2c)}{1} = 0$

$\implies a+b = 2c$

$\implies \frac{a+b}{2} = c \:---(1)$

$\alpha \beta = \frac{C}{A}\\= \frac{ab-(a+b)c}{1} \\= ab - (a+b) \frac{(a+b)}{2} \: [From \:(1)]$

$= \frac{2ab - (a+b)^{2}}{2}\\= \frac{2ab- (a^{2}+2ab+b^{2}}{2}\\= \frac{ 2ab - a^{2}-2ab -b^{2}}{2}\\= \frac{-a^{2}-b^{2}}{2} \\= \frac{-(a^{2}+b^{2})}{2}$

Therefore.,

$\red {\alpha \beta } \green {= \frac{-(a^{2}+b^{2})}{2}}$

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Answer 2

Answer:

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