Question

(x)2 +4x +k=0 the equation has real root when​

Answer 1

Step-by-step explanation:

x^2+4x+k=0

D=b^2-4ac

=(4)^2-(4)(1)(k)

=16-4k

D=0(Because roots are real)

16-4k=0

4-k=0

k=4

Answer 2

From the equation

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

What do we see from here?

Concept

• Complete the Square
• Quadratic Discriminant

Before we start

We might see them differently, but they have the same meaning.

Let's take a view from the quadratic formula.

→ $\displaystyle{\sf{(x^2+\frac{b}{a} x+\frac{b^2}{4a^2} )=\frac{b^2-4ac}{4a^2} }}$

→ $\displaystyle{\sf{(x+\frac{b}{2a} )^2=\frac{b^2-4ac}{4a^2} }}$

We get different solutions according to the right-hand side.

And this is the fundamental of the discriminant.

Solving the Problem

Only when $\sf{b^2-4ac}$ is negative, we get imaginary solutions.

We need $\sf{b^2-4ac\geq 0}$

• Discriminant $\sf{(4)^2-4\times(1)\times{k}}$
• We obtain $\sf{-4k+16\geq 0}$

Answer

We obtain real roots when $\sf{k\leq 4}$.