Question

the positive value of k for which the equation x2+kx+64 =0 has real roots is...​

Answer 1

you just need to know the meaning or equal and unequal roots...!

Answer 2

$p(x)=x^2+kx+64=0\\ \\ \\ a=1\ \ \ ; \ \ \ b=k\ \ \ ; \ \ \ c=64$

If the roots of  p(x)  are real, then the discriminant should be greater than or equal to 0.

We know,

→  If discriminant is greater than 0, then  p(x)  will have two real roots.

→  If discriminant is equal to 0, then  p(x)  will have only one real root.

→  If discriminant is lesser than 0, then  p(x)  will have no real roots.

Hence the first two possibilities can either be true.

So,

$\begin{aligned}&b^2-4ac\geq 0\\ \\ \Longrightarrow\ \ &k^2-(4\times 1\times 64)\geq 0\\ \\ \Longrightarrow\ \ &k^2-256\geq 0\\ \\ \Longrightarrow\ \ &k^2\geq 256\\ \\ \Longrightarrow\ \ &\sqrt{k^2}\geq \sqrt{256}\\ \\ \Longrightarrow\ \ &|k|\geq 16\end{aligned}$

But since k has positive value,  $|k|=k.$  So,

$\Longrightarrow\ \ \large \bold{k\geq 16}$

Or we can say,

$\large \bold{k\in [16, +\infty)}$