Question

• Class - 10 •

If α and β are the zeros of the quadratic polynomial f(x) = kx² + 4x + 4 such that α² + β² = 24, find the values of k.

Answer 1

Given that, α and β are the zeros of the Quadratic Polynomial f(x) = kx² + 4x + 4. And, (α² + β²) = 24.

As we know that,

Sum and Product of any Quadratic Polynomial are given by :

• (α + β) = (–b)/a
• (α β) = c/a

So, Sum and Product of given Quadratic polynomial kx² + 4x + 4 are,

• Sum of zeroes (α + β) = (–4)/k
• Product of zeroes (α β) = 4/k

$\rule{100px}{.3ex}$

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⭒By using I D E N T I T Y :

$\bigstar\;{\underline{\boxed{\pink{\frak{ \pmb{(a + b)^2 = a^2 + b^2 + 2ab}}}}}}$

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Therefore,

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$\sf :\implies \Big(\alpha + \beta \Big)^2 = \alpha^2 + \beta^2 + 2\alpha \beta \\\\\\:\implies\sf \bigg(\dfrac{-4}{\;k}\bigg)^2 = 24 + 2 \bigg(\dfrac{4}{k}\bigg) \\\\\\:\implies\sf \bigg(\dfrac{4^2}{k^2} \bigg)= 24 + 2 \bigg(\dfrac{4}{k}\bigg) \\\\\\:\implies\sf \bigg(\dfrac{16}{k^2}\bigg) = 24 + 8k \\\\\\:\implies\sf 16 = 24k^2 + 8k \\\\\\:\implies\sf 2 = 3k^2 + k \\\\\\:\implies\sf 3k^2 + k -2 = 0 \\\\\\:\implies\sf 3k^2 + 3k - 2k - 2 = 0 \\\\\\:\implies\sf 3k(k + 1) -2(k + 1) = 0 \\\\\\:\implies\sf (3k - 2) (k + 1) = 0 \\\\\\:\implies\underline{\boxed{\frak{\pmb{\pink{k = \dfrac{2}{3} \:or\; -1}}}}}\;\bigstar$

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$\therefore{\underline{\sf{Hence,\; the\;required\:value\;of\;k\;are\; \sf\pmb{ \dfrac{2}{3},\;-1.}}}}$

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Answer 2

Answer:

Given :-

If α and β are the zeros of the quadratic polynomial f(x) = kx² + 4x + 4 such that α² + β² = 24,

To Find :-

Value of k

Solution :-

Sum of zeroes

$\pmb { \alpha + \beta = \dfrac{ - b}{a}}$

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

Product of zeroes

$\pmb{ \alpha \beta = \dfrac{c}{a}}$

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

Apply Identity

${ \bigg( \alpha + \beta \bigg)}^{2} = { \alpha }^{2} + 2 \alpha \beta + { \beta }^{2}$

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

$\sf\dfrac{16}{ {k}^{2} } = 24 + 8k$

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

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

$\sf \: 2 = 24 {k}^{2} + k$

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

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

$\sf \: (3k - 2) \: or \: (k + 1)$

Either

k = 2/3

Or

k = -1