Question

If ÀLFA and BETA are the zeros of polynomial P(x) kx (sq) +4x+4 such that Alfa (sq)+Beta (sq) =24​

Answer 1

GIVEN :–

• A quadratic equation p(x) = kx² + 4x + 4 have two roots α and β .

• α² + β² = 24.

TO FIND :–

• Value of 'k' = ?

SOLUTION :–

• We know that –

$\\ \longrightarrow \sf \: \: sum \: \: of \: \: roots = - \dfrac{coffieciant \: \: of \: x}{coffieciant \: \: of \: \: {x}^{2} }$

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

$\\ \longrightarrow \sf \: \: product \: \: of \: \: roots = \dfrac{constant \: \: term}{coffieciant \: \: of \: \: {x}^{2} }$

$\\ \implies \sf \: \: \alpha \beta = \dfrac{4}{k }$

• Using given condition –

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

• Using identity –

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

• So that –

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

• Put the values –

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

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

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

$\\ \implies \sf \: \: 24 {k}^{2} + 8k - 16 = 0$

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

• Splitting Middle term –

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

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

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

$\\ \implies \large{ \boxed{ \sf \: k = - 1 \: , \: \dfrac{2}{3} }}$

Answer 2

$\star {\tt{\purple{\underline{Given:- }}}}$

★ ${\tt{\red{A\:,quadratic \: equation\:p(x) = kx^2 + 4x +4 \: have \: two \: roots α \: and \: β}}}$

${\implies{\tt{\red{α ^2 + β^2 = 24}}}}$

$\star {\tt{\purple{\underline{To \: Find:- }}}}$

★ ${\tt{\pink{value\: of \:k = ?}}}$

$\star {\tt{\purple{\underline{SoluTion :- }}}}$

${\mapsto{\tt{sum \: of \: the \: roots =\frac{cofficient \: of \: x}{cofficient\: of \: x^2}}}}$

${\implies{\tt{α + β = \frac{-4}{k}}}}$

${\mapsto{\tt{product \: of \: roots = \frac{constant \: term}{cofficient \: of \: x^2}}}}$

${\implies{\tt{αβ = \frac{4}{k}}}}$

★ ${\tt{\pink{\:Using\: the \:given}}}$

${\implies{\tt{\blue{α ^2 + β^2 = 24}}}}$

★ ${\tt{\pink{ Formula\: used}}}$

${\tt{\underline{(a + b)^2 = a^2 +b^2 +2ab}}}$

★ ${\tt{\pink{Using\: the \: Formula \: we \: get, }}}$

${\implies{\tt{(α + β) ^2 - 2αβ= 24}}}$

★ ${\tt{\pink{putting\: the \: values\: we \: get, }}}$

${\implies{\tt{(\frac{-4}{k}^2) - 2 (\frac{4}{k}) = 24}}}$

${\implies{\tt{(\frac{16}{k}^2) - 2 (\frac{8}{k}) = 24}}}$

${\implies{\tt{16 - 8k = 24k^2}}}$

${\implies{\tt{24k^2 +8k - 16 = 0}}}$

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

★ NoW, using the method of middle term,

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

${\implies{\tt{3k(k+1) - 2 (k +1)= 0}}}$

${\implies{\tt{(3k- 2)(k+1)= 0}}}$

$\implies\boxed {\tt{ \: K =1 , \frac{2}{3}}}$