Question

9. If ∝ and are two zeroes of the polynomial kx

2 + 4 x + 4 such that
∝
2 +
= 24 , then find the value of k .
plzz ans it fast .i will u brainliest​

Answer 1

Step-by-step explanation:

According to the question,

α and β are the zeroes of polynomial kx² + 4x + 4

a = k

-b = -4

c = 4

α + β = -b/a = -4/k

αβ = c/a = 4/k

Given => α² + β² = 24

=> (α + β)² - 2αβ = 24

=> (-4/k)² - 2×4/k = 24

=> 16/k² - 8/k = 24

=> 16 - 8k = 24k²

=> 24k² + 8k - 16 = 0

=> 3k² + k - 2 = 0

=> 3k² + 3k - 2k - 2 = 0

=> 3k(k + 1) - 2(k + 1) = 0

=> (3k - 2)(k + 1) = 0

=> k = 2/3 or -1

Answer 2

${\green {\boxed {\mathtt {✓verified\:answer}}}}$

$since \: \alpha \: and \: \beta \: are \: the \: zeros \: of \: polynomial \: p(x) = k {x} + 4x + 4 \: then \\ \alpha + \beta = \frac{ - b}{a} = \frac{ - 4}{k} \: and \\ \alpha \beta = \frac{c}{a} = \frac{4}{k} \: \\ it \: is \: given \: that \: { \alpha }^{2} + { \beta }^{2} = 24 \\ \therefore( \alpha + \beta )^{2} - 2 \alpha \beta = 24 \\ ( {\frac{ - 4}{k} })^{2} - 2 ({\frac{4}{k} }) {}^{2} = 24 \\ \frac{16}{{k} ^{2} } - \frac{8}{k} = 24 \\ \frac{16 - 8k}{k {}^{2} } = 24 \\ 16 - 8k = 24 {k}^{2} \\ 24 {k}^{2} + 8k - 16 = 0 \\ 8(3 {k}^{2} + k - 16) = 0 \\ 3k {}^{2} + k - 2 = 0 \\ 3k {}^{2} + 3k - 2k - 2 = 0 \\ 3k(k + 1) - 2(k + 1) = 0 \\ (k + 1) = 0 \: or \:( 3k - 2 )= 0 \\ \therefore \: k = - 1 \: or \: k = \frac{2}{3}$

hope it helps you....