Value of roots

Algebra Level 3

If α , β \alpha , \beta be the roots of x 2 − a ( x − 1 ) + b = 0 x^{2} -a(x - 1) + b = 0 , then the value of 1 ( α ) 2 − a α + 1 ( β ) 2 − a β + 2 a + b \frac{1}{ (\alpha)^{2} - a\alpha } + \frac{1}{ (\beta)^{2} - a\beta } + \frac{2}{a + b} is


The answer is 0.

Sakanksha Deo by Sakanksha Deo , 23, India

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1 solution

Sakanksha Deo
Mar 8, 2015

So we have,

x 2 − a ( x − 1 ) + b = 0 x^{2} - a(x - 1) + b = 0

⇒ x 2 − a x + a + b = 0 \Rightarrow x^{2} -ax +a + b = 0

Now,we know that

Product of roots = α β = c o s n t a n t t e r m = a + b \alpha\beta = cosntant term = a + b ......... (1)

s u m o f r o o t s = α + β = a sum of roots = \alpha + \beta = a ....... (2)

Now

1 ( α ) 2 − a α + 1 ( β ) 2 − a β + 2 a + b \frac{1}{ (\alpha)^{2} - a\alpha } + \frac{1}{ (\beta)^{2} - a\beta } + \frac{2}{a + b}

= 1 α ( α − a ) + 1 β ( β − a ) + 2 a + b = \frac{1}{ \alpha(\alpha - a) } + \frac{1}{ \beta (\beta - a) } + \frac{2}{a + b}

Now,from (1) and (2),

= 1 ( α ) ( − β ) + 1 ( β ) ( − α ) + 2 α β = \frac{1}{ (\alpha)(-\beta) } + \frac{1}{ (\beta)(-\alpha) } + \frac{2}{\alpha\beta}

= − 1 α β + − 1 α β + 2 α β = 0 = \frac{-1}{\alpha\beta} + \frac{-1}{\alpha\beta } + \frac{2}{\alpha\beta} = \boxed{0}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Better if you give it as multiple choice type.

Sai Ram - 5 years, 11 months ago

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Can't change it now

Sakanksha Deo - 5 years, 11 months ago

i think α + β = − a \alpha +\beta = -a

Rishabh Tripathi - 6 years, 2 months ago

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