Question

If α and β are the zeroes of the polynomial f(x)= x2
-6x +k,
find the value of k, such that α
^2+b^2 = 40.​

Answer 1

$p(x) = {x}^{2} - 6x + k$

$let \: p(x) = 0$

${x}^{2} - 6x + k = 0$

$now \: { \alpha }^{2} + { \beta }^{2} = 40$

$as \: we \: know \: that \: \alpha = - \frac{b}{a} and \: \beta = \frac{c}{a}$

$a = 1$

$b = - 6$

$c = k$

$so \: { \alpha }^{2} + { \beta }^{2} = 40$

$(\frac{ - b}{a})^{2} \: + ( \frac{c}{a}) {}^{2} = 40$

$( \frac{ - ( - 6)}{1}) {}^{2} + ( \frac{k}{1}) {}^{2} = 40$

${6}^{2} + {k}^{2} = 40$

$36 + k^{2} = 40$

$k^{2}= 4$

k=+or-2