Question

For what value of k the root of the equation is ${{x}^{2}}+4x+k=0$ are real?

Answer 1

For every quadratic equation whose roots are real, the discriminant D =$\sqrt{{{b}^{2}}-4ac}$ of the given equation must be greater and equal to zero i.e. $\sqrt{{{b}^{2}}-4ac}\ge 0$
where a = coefficient of ${{x}^{2}}$
b = coefficient of $x$
c = constant

Complete step-by-step solution:
Let us take $a{{x}^{2}}\text{+ }bx+c=0$ to be a quadratic equation.
We know that the roots of the above equation are given by:
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
where, $\sqrt{{{b}^{2}}-4ac}$ is known as discriminant(D).
Now, comparing given equation ${{x}^{2}}+4x+k=0$ with general equation $a{{x}^{2}}\text{+ }bx+c=0$, we will get:
a = 1, b = 4, and c = k.
Now, we will use formula to find the discriminant of equation ${{x}^{2}}+4x+k=0$ given by:
D = $\sqrt{{{b}^{2}}-4ac}$ --------(1)
Here, we have a = 1, b = 4, and c = k
Putting the value of a, b, c in equation (1), we get
$\Rightarrow D=\sqrt{{{(4)}^{2}}-4(1)(k)}$
$\Rightarrow D=\sqrt{16-4k}$
Since, the roots of the above given equation in question i.e. ${{x}^{2}}+4x+k=0$ is real. Hence, $D\ge 0$
$\therefore \sqrt{16-4k}\ge 0$
Squaring both side of the above inequality, we will get:
$\Rightarrow \left( 16-4k \right)\ge 0$
Taking $4$ common from the Left-hand side of the above inequality, we will get:
$\Rightarrow 4\left( 4-k \right)\ge 0$
It can also be written as:
$\Rightarrow \left( 4-k \right)\ge 0$
Take $k$ from Left-hand to Right-hand side, we will get: -
$\Rightarrow 4\ge k$
$\Rightarrow k\le 4$
Hence, k must be less than 4 for the root of the equation ${{x}^{2}}+4x+k=0$ to be real.
Hence, $k\le 4$ is the answer.

Note: Two equations can only be compared if and only if the Right-hand side of both the equations is zero, otherwise there is always the chance of error. And, also comparison can only be done between variables of the same degree. Squaring both sides of inequality can only be done if the Right-hand side of the inequality is positive constant.