Question

For what values of K does x^2+kx+4=0 have two equal real roots?

Answer 1

Answer:

For k= +4, the polynomial has two equal roots.

For k  = -4, the polynomial has two equal roots.

Step-by-step explanation:

Here, the given expression is :$x^{2} + kx + 4 = 0$

(1) Now, let us assume the value of k = +4

So, the expression is $x^{2} + 4x + 4 = 0$

Simplifying the given by splitting,

$x^{2} + 4x + 4 = 0 \implies x^{2} + 2 x + 2x + 4\\\implies x(x+2) +2(x+2) =0\\or, (x+2)(x+2) = 0$

$x = -2, x= -2$

Hence, for k= +4, the polynomial has two equal roots.

Now, let us assume the value of k = -4

So, the expression is $x^{2} - 4x + 4 = 0$

Simplifying the given by splitting,

$x^{2} - 4x + 4 = 0 \implies x^{2} - 2 x - 2x + 4\\\implies x(x-2) -2(x-2) =0\\or, (x-2)(x-2) = 0$

$x = 2, x= 2$

Hence, for k  = -4, the polynomial has two equal roots.

(3) Let us assume the value of k < -4

This gives us NO FIXED VALUE for k.

So, the expression is $x^{2} + 4x + 4 = 0$ can not be solved.

(4)  let us assume the value of k > 4

This gives us NO FIXED VALUE for k.

So, the expression is $x^{2} + 4x + 4 = 0$ can not be solved.

Hence, for k =+4, and k = -4 the polynomial has two equal roots.