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Question: If a and b are the roots of the equation x<sup>2</sup> – p(x + 1) – q = 0, then the value of \(\fra...

If a and b are the roots of the equation

x2 – p(x + 1) – q = 0, then the value of α2+2α+1α2+2α+q\frac{\alpha^{2} + 2\alpha + 1}{\alpha^{2} + 2\alpha + q}+

β2+2β+1β2+2β+q\frac{\beta^{2} + 2\beta + 1}{\beta^{2} + 2\beta + q}is –

A

2

B

3

C

0

D

1

Answer

1

Explanation

Solution

We have x2 – px – (p + q) = 0

\ a + b = p and ab = – (p + q)

\ α2+2α+1α2+2α+q\frac{\alpha^{2} + 2\alpha + 1}{\alpha^{2} + 2\alpha + q}+ β2+2β+1β2+2β+q\frac{\beta^{2} + 2\beta + 1}{\beta^{2} + 2\beta + q}

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

=2(α+1)2(β+1)2+(q−1)[(α+1)2+(β+1)2](α+1)2(β+1)2+(q−1)[(α+1)2+(β+1)2]+(q−1)2\frac{2(\alpha + 1)^{2}(\beta + 1)^{2} + (q - 1)\lbrack(\alpha + 1)^{2} + (\beta + 1)^{2}\rbrack}{(\alpha + 1)^{2}(\beta + 1)^{2} + (q - 1)\lbrack(\alpha + 1)^{2} + (\beta + 1)^{2}\rbrack + (q - 1)^{2}}

Now (a + 1) (b + 1) = ab + a + b + 1

= – (p + q) + p + 1 = 1 – q

\ Given expression

= 2(1−q)2+(q−1)[(α+1)2+(β+1)2](1−q)2+(q−1)[(α+1)2+(β+1)2]+(1−q)2\frac{2(1 - q)^{2} + (q - 1)\lbrack(\alpha + 1)^{2} + (\beta + 1)^{2}\rbrack}{(1 - q)^{2} + (q - 1)\lbrack(\alpha + 1)^{2} + (\beta + 1)^{2}\rbrack + (1 - q)^{2}}= 1