Question

solve these ques. guys .. n give me solution as soon as possible : )

Answer 1

Hello Dear !
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1) here equation => a (x²+1) - x(a² + 1 ) = 0
Quadratic polynomial = a (x²+1) - x(a² + 1 )
                                   = ax² + a - a²x - x
                                   = ax² - x -a²x + a [ax² -(1+a²)x + a )]
                                   = x(ax-1) -a(ax-1)
                                   = (ax-1) (x-a) 
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So, zeros are x = 1/a and x = a 
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Relation between coefficients :

-b/a = 1/a + a
-(-1 - a²) /a = 1+a² /a
1+a²/a        =  1+a² /a
LHS = RHS 

c/a = 1/a x a 
a/a = a/a 
1    = 1 
LHS = RHS
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2) α and β are two zeros of quadratic equation 6x² - x + 1 
    Find  $\frac{a}{b} + \frac{b}{a} + 2( \frac{1}{a}+ \frac{1}{b} )$
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we know α + β = -b/a = 1/6 
                αβ    = c/a   = 1/6 
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= $\frac{a}{b} + \frac{b}{a} + 2( \frac{1}{a}+ \frac{1}{b} )$

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

= $\frac{11}{6} +2$$\frac{(a+b)^2 - 2ab}{ab} + 2( \frac{a+b}{ab} )$

= $\frac{(1/6)^2 - 2 /6 }{1/6} + 2[(1/6) /(1/6) ]$

= $\frac{ \frac{1}{36}- \frac{1}{3} }{1/6} + 2$

= $\frac{ \frac{1-12}{36} }{1/6} + 2$

= $\frac{11/36}{1/6} +2$

=$\frac{11+12}{6}$

=$\frac{23}{6}$