Question

If alpha and bita are two zeros of the polynomials kx^2+4x+4 such that alpha^2+bita^2=24,then find the value of k

Answer 1

Answer:

Step-by-step explanation:

Given,

quadratic equation k x²+4 x+4

α+β= -b/a

      =-4/k

αβ= c/a

    =4/k

but given

α²+β² = 24

we know that

a²+b²=(a+b)²-2ab

⇒(α+β)²-2αβ=24

⇒(-4/k)²-2×4/k=24

⇒16/k² -8/k = 24

⇒16-8 k/k² = 24

⇒16-8 k = 24 k²

⇒24 k²+8 k - 16=0

⇒8(3 k²+k-2)=0

⇒3 k²+3 k-2 k-2=0

⇒3 k(k+1)-2(k+1)=0

⇒(k+1)(3 k-2)=0

⇒k+1=0     (or)     3 k-2=0

⇒k= -1        (or)      k=2/3

∴k= -1 (or) 2/3

hope you understand

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Answer 2

$\star \;\;\; {\underline {\underline {\mathfrak {\pink {AnsWeR:-}}}}}$

• $\alpha$ and $\beta$ are two zeros of the polynomials kx² +4x +4

• $\sf { \alpha}^2 + { \beta}^2 = 24$

$\star \;\;\; {\underline {\underline {\mathfrak {\pink {To \; Find:-}}}}}$

• Value of k.

$\star \;\;\; {\underline {\underline {\mathfrak {\green {Solution:-}}}}}$

$\alpha$ and $\beta$ are two zeros of the polynomials kx² +4x +4 = 0

We know that,

$\dag\;{\underline{\underline{\boxed{\sf{\purple{Sum\;of\;zeroes\;( \alpha + \beta) = \dfrac{-b}{a}}}}}}} \\\\ :\implies\sf ( \alpha + \beta) = \dfrac{- 4}{k}$

$\dag\;{\underline{\underline{\boxed{\sf{\purple{Product;of\;zeroes\;( \alpha \beta) = \dfrac{c}{a}}}}}}} \\\\ :\implies\sf ( \alpha + \beta) = \dfrac{4}{k}$

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★ According to question:-

$\star\;\;\;\sf { \alpha}^2 + { \beta}^2 = 24$

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

$\;\;\;\;\star\;\;\normalsize{\underline{\underline{\sf{\red{Putting\;value\;( \alpha + \beta)\;and\;( \alpha \beta)\;:-}}}}}$

$:\implies\sf \bigg( \dfrac{-4}{k} \bigg)^2 - 2 \bigg( \dfrac{4}{k} \bigg) = 24 \\\\ :\implies\sf \dfrac{16}{k^2} - \dfrac{-8}{k} = 24 \\\\ :\implies\sf 8 \bigg( \dfrac{2}{k^2} - \dfrac{-1}{k} - 3 \bigg) = 0 \\\\ :\implies\sf 2 - k = 3k^2 \\\\ :\implies\sf 3k^2 + k - 2 = 0 \\\\ :\implies\sf 3k^2 + 3k - 2k - 2 = 0$

$:\implies\sf 3k(k + 1) -2(k + 1) = 0 \\\\ :\implies\sf (3k - 2)(k + 1) = 0 \\\\ :\implies\sf Either\;(3k - 2)\;and\;(k + 1)\;is\;equals\;to\;zero \\\\ :\implies\sf Hence,\; k = \dfrac{2}{3} \;and -1$

${\underline{\underline{\sf{\pink{\dag\;Hence\;Solved!!}}}}}$

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