Question

if alpha and beta are zeroes of a polynomial kx square + 4x + 4 such that alpha square + beta square = 24, find the value of K ??​

Answer 1

$p(x) = kx^2+4x+4=0\\ \\ \\ a=k\ \ \ ; \ \ \ b=4\ \ \ ; \ \ \ c=4$

Zeroes of the polynomial are α and β.

$\displaystyle \alpha+\beta=-\frac{b}{a} \\ \\ \\ \alpha+\beta=-\frac{4}{k}$

$\displaystyle \alpha\beta=\frac{c}{a} \\ \\ \\ \alpha\beta=\frac{4}{k}$

Given that,

$\alpha^2+\beta^2=24$

$\displaystyle (\alpha+\beta)^2=\left(-\frac{4}{k}\right)^2 \\ \\ \\ \alpha^2+\beta^2+2\alpha\beta=\frac{16}{k^2} \\ \\ \\ 24+2 \cdot \frac{4}{k}=\frac{16}{k^2}\ \ \ \ \ \ \ \ \ \ \left[\because\ \alpha^2+\beta^2=24\ \ \ ; \ \ \ \alpha\beta=\frac{4}{k}\right] \\ \\ \\ 24+\frac{8}{k}=\frac{16}{k^2} \\ \\ \\ \frac{16}{k^2}-\frac{8}{k}=24 \\ \\ \\ \frac{16}{k^2}-\frac{8k}{k^2}=24 \\ \\ \\ \frac{16-8k}{k^2}=24 \\ \\ \\ 16-8k=24k^2 \\ \\ \\ 8(2-k)=8 \cdot 3k^2 \\ \\ \\ 2-k=3k^2 \\ \\ \\ 3k^2+k-2=0$

$\displaystyle 3k^2+3k-2k-2=0 \\ \\ \\ 3k(k+1)-2(k+1)=0 \\ \\ \\ (k+1)(3k-2)=0 \\ \\ \\ \\ \therefore\ k=\bold{-1}\ \ \ ; \ \ \ k=\bold{\frac{2}{3}}$

Thus the value of k is either -1 or 2/3.

Answer 2

Answer:

Some error in your question , question is in such a way --> α and β are the zero of the Kx² + 4x + 4 , α² + β² = 24 then find k ?

Solution :- α and β are the zeros of the given polynomial Kx² + 4x + 4 = 0

so, product of zeros = αβ = constant/coefficient of x² = 4/K

sum of zeros = α + β = -coefficient of x/Coefficient of x² = -4/k

Now, α² + β² = 24

⇒(α + β)² - 2αβ = 24

⇒(-4/k)² - 2(4/k) = 24

⇒16/K² - 8/k = 24

⇒ 2 - k = 3k²

⇒3k² + k -2 = 0

⇒ 3k² + 3k - 2k - 2 = 0

⇒3k(k + 1) - 2(k +1) = 0

⇒(3k -2)(k + 1) = 0

Hence, k = 2/3 and -1

hope help u mate ✌