Question

if alfa and beta are zeroes of the quadratic polynomial p(x)=kx^2+x_6 such that alfa^2+beta^2=25/4, find the value of k

Answer 1

Hi Mate !!

K = 2 and k = -2/25 will satisfy the equation p(x)

Answer 2

$given - - > p(x) = k {x}^{2} + x - 6 \\ { \alpha }^{2} + { \beta }^{2} = \frac{25}{4} \\ \\ by \: given \\ \\ a = k \\ b = 1 \\ c = - 6 \\ \\ \alpha + \beta = \frac{ - b}{a} \\ = - \frac{1}{k} \\ \alpha \beta = \frac{c}{a} \\ = - \frac{6}{k} \\ \\ { \alpha }^{2} + { \beta }^{2} = \frac{25}{4} \\ {( \alpha + \beta )}^{2} - 2 \alpha \beta = {( \frac{5}{2} )}^{2} \\ ( - { \frac{1}{k}) }^{2} - 2( - \frac{6}{k} ) = \frac{25}{4} \\ \frac{1}{ {k}^{2} } + \frac{12}{k} = \frac{25}{4} \\ 4(1 + 12k) = 25 {k}^{2} \\ 4 + 48k = 25 {k}^{2} \\ 25 {k}^{2} - 48k - 4 = 0 \\ 25 {k}^{2} - 50k + 2k - 4 = 0 \\ 25k(k - 2) + 2(k - 2) = 0 \\ (k - 2)(25k + 2) = 0 \\ k = 2 \: and \: - \frac{2}{25}$