Question

If alpha and beta are zeros of p(X)= x^2 -6x+k . Find the value of k such that (alpha + beta)^2-2alpha beta =40

Answer 1

Answer:

2

Step-by-step explanation:

Given that $\alpha, \beta$ are the roots of the equation $p(x) = x^2 - 6x + k$.

Also, $(\alpha + \beta)^2 - 2 \alpha\beta = 40...(1)$

From the theory of quadratic equations, we know that:

$\text{sum of roots} = \alpha + \beta = \dfrac{-\text{coefficient of } x}{\text{coefficient of } x^2} = \dfrac{-(-6)}{1} = 6$

$\text{product of roots} = \alpha\beta = \dfrac{\text{constant term}}{\text{coefficient of } x^2} = \dfrac{k}{1} = k$

Substituting in equation (1),

$6^2 -2k = 40\\36 - 2k = 40\\2k = 4\\k = 2$

Therefore, the value of k is 2.

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