Question

if alpha and beta are the zeroes of the quadratic polynomial f{x}=kx^2+4x+4 such that alpha^2+beta^2=24,find the value of k.

Answer 1

Given :-

• $\sf \: \alpha \: and \: \beta \: are \: zeroes \: of \green{\: k {x}^{2} + 4x + 4 }$

To Find :-

• Value of K

Solution :-

$\dag\underline{\sf\green{{\sf Sum\:of\:Zeros:- }}}$

$:\implies\sf \alpha+\beta=\dfrac{-b}{a}$

$\sf :\implies\: \alpha + \beta = - \: \dfrac{4}{k} - - - (1)$

$\dag\underline{\sf\green{{\sf Product\:of\:Zeros :-}}}$

$\sf:\implies\: \alpha \beta \: =\: \dfrac{4}{k} - - - (2)$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

⠀⠀⠀⠀⠀$\small\underline{\pmb{\sf According \: to \: the \: question :-}}$

$\sf \: \: \: \: \: \: \: \: \: \:\pink{ :\implies { \alpha }^{2} + { \beta }^{2} = 24}$

$\sf \: \: \: \: \: \: \: \: \: \:\: :\implies{(\alpha + \beta ) }^{2} - 2 \alpha \beta = 24$

$\sf \: \: \: \: \: \: \: \: \: \:\::\implies {\bigg( - \dfrac{4}{k} \bigg) }^{2} - 2 \times \dfrac{4}{k} = 24$

$\sf \: \: \: \: \: \: \: \: \: \:\::\implies\dfrac{16}{ {k}^{2} } - \dfrac{8}{k} = 24$

$\sf\: \: \: \: \: \: \: \: \: \: \::\implies\dfrac{16 - 8k}{ {k}^{2} } = 24$

$\sf \: \: \: \: \: \: \: \: \: \:\::\implies2 - k = {3k}^{2}$

$\sf \: \: \: \: \: \: \: \: \: \:\: :\implies{3k}^{2} + k - 2 = 0$

$\sf \: \: \: \: \: \: \: \: \: \:\: :\implies{3k}^{2} + 3k - 2k - 2 = 0$

$\sf \: \: \: \: \: \: \: \: \: \:\::\implies3k(k + 1) - 2(k + 1) = 0$

$\sf \: \: \: \: \: \: \: \: \: \:\::\implies(k + 1)(3k - 2) = 0$

$\sf \pink{\sf \: \: \: \: \: \: \: \: \: \::\implies\:k = - 1 \: \: \: or \: \: \: k = \dfrac{2}{3}}$

$\therefore\:\underline{\textsf{ Value of K is \textbf{-1 \: Or \: 2/3}}}.$