Question

if alpha and beta are the zeros of the quadratic polynomial f x is equal to K X square + 4 x + 4 such that Alpha square plus beta square is equal to 24 find the value of k​

Answer 1

It's fundamental question of quadratic equation.

Answer 2

Answer:

The value of k is

$\boxed{\red{\sf\:k\:=\:-\:1}}\:\:\:\:\sf\:or\:\:\:\boxed{\red{\sf\:k\:=\:\frac{2}{3}}}$

Step-by-step-explanation:

The given quadratic equation is $\sf\:kx^{2}\:+\:4x\:+\:4\:=\:0$

We have given that, $\sf\:\alpha\:\&\:\beta$ are the roots of the given quadratic equation such that $\sf\:\alpha^{2}\:+\:\beta^{2}\:=\:24$

We know that,

$\sf\:kx^{2}\:+\:4x\:+\:4\:=\:0$

Comparing with $\sf\:ax^{2}\:+\:bx\:+\:c\:=\:0$, we get,

$\sf\:a\:=\:k\\\\\sf\:b\:=\:4\\\\\sf\:c\:=\:4$

We know that,

$\sf\:Sum\:of\:roots\:=\:\alpha\:+\:\beta\:=\:-\:\frac{b}{a}\\\\\implies\sf\:\alpha\:+\:\beta\:=\:-\:\frac{4}{k}\\\\\sf\:Now,\\\\\sf\:Product\:of\:roots\:=\:\alpha\:.\:\beta\:=\:\frac{c}{a}\\\\\implies\sf\:\alpha\:.\:\beta\:=\:\frac{4}{k}$

Now, we know that,

$\sf\:\alpha^{2}\:+\:\beta^{2}\:=\:(\:\alpha\:+\:\beta\:)^{2}\:-\:2\:\alpha\:.\:\beta\:\:\:\:-\:-\:[\:Identity\:]\\\\\implies\sf\:\alpha^{2}\:+\:\beta^{2}\:=\:(\:-\:\frac{4}{k}\:)^{2}\:-\:2\:\times\:\frac{4}{k}\\\\\implies\sf\:\alpha^{2}\:+\:\beta^{2}\:=\:\frac{16}{k^{2}}\:-\:\frac{8}{k}\\\\\implies\sf\:\frac{16}{k^{2}}\:-\:\frac{8k}{k^{2}}\\\\\implies\sf\:\dfrac{16\:-\:8k\:}{k^{2}}$

But,

$\sf\:\alpha^{2}\:+\:\beta^{2}\:=\:\:\:-\:-\:-\:-[\:Given\:]\\\\\therefore\sf\:\dfrac{16\:-\:8k}{k^{2}}\:=\:24\\\\\implies\sf\:16\:-\:8k\:=\:24\:k^{2}\\\\\implies\sf\:24k^{2}\:+\:8k\:-\:16\:=\:0\\\\\implies\sf\:3k^{2}\:+\:k\:-\:2\:=\:0\:\:\:-\:-\:[\:Dividing\:both\:sides\:by\:8\:]\\\\\implies\sf\:3k^{2}\:+\:3k\:-\:2k\:-\:2\:=\:0\\\\\implies\sf\:3k\:(\:k\:+\:1\:)\:-\:2\:(\:k\:+\:1\:)\:=\:0\\\\\implies\sf\:(\:k\:+\:1\:)\:(\:3k\:-\:2\:)\:=\:0\\\\\implies\sf\:k\:+\:1\:=\:0\:\:\:or\:\:\:3k\:-\:2\:=\:0\\\\\implies\sf\:k\:=\:-\:1\:\:\:or\:\:\:3k\:=\:2\\\\\implies\boxed{\red{\sf\:k\:=\:-\:1}}\:\:\:\:\sf\:or\:\:\:\boxed{\red{\sf\:k\:=\:\frac{2}{3}}}$