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Question: If the roots of the equation \(x^{2} - 5x + 16 = 0\) are α, β and the roots of equation \(x^{2} + px...

If the roots of the equation x2−5x+16=0x^{2} - 5x + 16 = 0 are α, β and the roots of equation x2+px+q=0x^{2} + px + q = 0 are α2+β2\alpha^{2} + \beta^{2}, αβ/2\alpha\beta/2, then

A

p=1,q=−56p = 1,q = - 56

B

p=−1,q=−56p = - 1,q = - 56

C

p=1,q=56p = 1,q = 56

D

p=−1,q=56p = - 1,q = 56

Answer

p=−1,q=−56p = - 1,q = - 56

Explanation

Solution

Since roots of the equation x2−5x+16=0x^{2} - 5x + 16 = 0 are α,β\alpha,\beta.

⇒ α+β=5,αβ=16\alpha + \beta = 5,\alpha\beta = 16 and α2+β2+αβ2=−p\alpha^{2} + \beta^{2} + \frac{\alpha\beta}{2} = - p

⇒(α+β)2−2αβ+αβ2=−p(\alpha + \beta)^{2} - 2\alpha\beta + \frac{\alpha\beta}{2} = - p ⇒ 25−2(16)+162=−p25 - 2(16) + \frac{16}{2} = - p ⇒ p=−1p = - 1

and (α2+β2)(αβ2)=q(\alpha^{2} + \beta^{2})\left( \frac{\alpha\beta}{2} \right) = q ⇒ [(α+β)2−2αβ]αβ2=q\lbrack(\alpha + \beta)^{2} - 2\alpha\beta\rbrack\frac{\alpha\beta}{2} = q

⇒ (25−32)8=q(25 - 32)8 = q ⇒ q=−56q = - 56