Question

if a and b are the zeros of the polynomial x^2 - 8x +k such that a^2 +b^2 =40 find the value of k​

Answer 1

$\Huge{\color{red}{\fbox{\textsf{\textbf{♠Solution ♠:-}}}}}$

Given :-

• p(x) → x² - 8x + k
• zeros → a , b

So,

★ Sum of zeros

$\bold{→ a + b = \frac{ - ( - 8)}{1} }$

$\boxed{\bold {→ a + b = 8}}$

★ Product of zeros ↓

$→ ab = \frac{k}{1}$

$\boxed {\bold{→ ab = k }}$

Again ,

$\bold{ \: \: \: \: {a}^{2} + {b}^{2} = 40}$

$\implies \: {(a + b)}^{2} - 2ab = 40$

Substituting the values we get ,

$\implies \: {8}^{2} - 2k = 40$

$\implies \: 64 - 2k = 40$

$\implies \: 2k = 64 - 40$

$\implies \: 2k = 24$

$\implies \: k = \frac{24}{2}$

$\implies \: \boxed{\bold{ k =12}}$

$\Large{\colorbox{yellow}{\underline{\underline{♠Answer♠:—\:\:k\:\:=\:12} }}}$