Question

Solve this question (◍•ᴗ•◍)❤​

Answer 1

Question :

If $\alpha$ and $\beta$ are the zeros of the polynomial P(x) = Kx² + 4x + 4 such that $\alpha^2+\beta^2$=24 , find the value of K.

Given :

• P(x) = Kx² + 4x + 4
• $\sf{\alpha^2+\beta^2=24}$

To find :

• Value of "K".

Solution :

$\sf{\alpha^2+\beta^2=24.............(1)}$

$\sf{\alpha\: and\:\beta\:are\:the\: zeros\:of\:the\: polynomial.}$

★ Compare Kx² + 4x +4 = 0 with ax² + bx + c = 0.

Where,

• a = K
• b = 4
• c = 4

We know,

• Sum of zeros = - b/a
• Product of zeros = c/a

Here,

★$\sf{Sum\:of\: zeros (\alpha+\beta)=\dfrac{-4}{K}}$

★$\sf{Product\:of\; zeros (\alpha\beta)=\dfrac{4}{K}}$

Now , take eq (1)

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

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

• Put values .

$\implies\sf{(\dfrac{-4}{K})^2-2\times\dfrac{4}{K}=24}$

$\implies\sf{\dfrac{16}{K^2}-\dfrac{8}{K}=24}$

$\implies\sf{\dfrac{16-8K^2}{K^2}=24}$

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

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

$\implies\sf{8(3K^2+K-2)=0}$

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

$\implies\sf{3K^2+(3-2)K-2=0}$

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

$\implies\sf{3K(K+1)-2(K+1)=0}$

$\implies\sf{(K+1)(3K-2)=0}$

Either,

K+1= 0

→ K = -1

Or,

3K - 2 = 0

→ 3K = 2

→ K = 2/3

Therefore, the value of K is -1 or 2/3.

Answer 2

† Given †

$\leadsto\bf\red{kx^2+4x+4}$

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

$\rule{150}2$

★ To find ★

→ value of "K"=?

we know,

$\leadsto\alpha \: and\: \beta$are the zeroes of the polynomial.

$\leadsto\alpha^2+\beta^2=24$--------eq(1)

On comparing $\leadsto kx^2+4x+4=0$ with $ax^{2}+bx+c=0$

so,

$\implies a=K\\ \implies b=4 \\ \implies c=4$

And,

Sum of the zeroes $\alpha+\beta=\frac{-b}{a}$

product of Zeroes $\alpha\beta=\frac{c}{a}$

$\leadsto\frac{-b}{a}=\frac{-4}{k}$

$\leadsto\frac{c}{a}=\frac{4}{k}$

$\rule{300}2$

$\longrightarrow \alpha^2+\beta^2=24----eq(1)$

$\longrightarrow (\alpha+\beta)-2\alpha\beta=24$

• using identity $(a+b)^2$↑

◇ According to question ◇

$\leadsto (\frac{-4}{k})^2-2×\frac{4}{k}=24\\ \leadsto \frac{16}{k^2}-\frac{8}{k}=24\\\leadsto\frac{16-8k^2}{k^2}=24\\ \leadsto 24k^2=16-8k\\ \leadsto 24k^2+8k-16=0\\ \leadsto 8(3k^2+k-2)=0\\ \leadsto 3k^2+k-2=0\\ \leadsto 3k^2+3k-2k-2=0\\ \leadsto 3k(k+1)-2(k+1)=0\\ \leadsto(k+1)(3k-2)=0\\ \leadsto k+1=0,3k-2=0\\ \leadsto k=-1, k=\frac{-2}{3}$

$\rule{150}2$

Hence,

$\huge\red{\fbox{\fbox{k=-1}}}$&$\huge\pink{\fbox{\fbox{k= -2/3} }}$