Question

How do you find all the real and complex roots of ${{x}^{2}}+8x+25=0$?

Answer 1

Hint: In this question we have a polynomial equation which has the degree $2$ therefore, this expression is in the form of a quadratic equation. We will solve the quadratic equation using the quadratic formula which is $({{x}_{1}},{{x}_{2}})=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ where $({{x}_{1}},{{x}_{2}})$ are the two roots of the equation and $a,b,c$ are the coefficients of the terms in the quadratic equation.

Complete step by step answer:
We have the quadratic equation given to us as ${{x}^{2}}+8x+25=0$
The general form of a quadratic equation is $a{{x}^{2}}+bx+c=0$ therefore, on comparing it with the given quadratic equation we can see the coefficients of the terms as:
$a=1$
$b=8$
$c=25$
On substituting the values in the quadratic formula $({{x}_{1}},{{x}_{2}})=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ we get:
$\Rightarrow ({{x}_{1}},{{x}_{2}})=\dfrac{-8\pm \sqrt{{{8}^{2}}-4(1)(25)}}{2(1)}$
On multiplying the terms and taking the square in the root part, we get:
\[\Rightarrow ({{x}_{1}},{{x}_{2}})=\dfrac{-8\pm \sqrt{64-100}}{2(1)}\]
On simplifying the denominator, we get:
\[\Rightarrow ({{x}_{1}},{{x}_{2}})=\dfrac{-8\pm \sqrt{64-100}}{2}\]
Now on simplifying the root part, we get:
\[\Rightarrow ({{x}_{1}},{{x}_{2}})=\dfrac{-8\pm \sqrt{-36}}{2}\]
Now the term $-36$ can be written as a product of $36$ and $-1$ therefore, on substituting, we get:
\[\Rightarrow ({{x}_{1}},{{x}_{2}})=\dfrac{-8\pm \sqrt{36\times -1}}{2}\]
Now we know that $\sqrt{36}=6$ therefore, on taking the square root, we get:
\[\Rightarrow ({{x}_{1}},{{x}_{2}})=\dfrac{-8\pm 6\sqrt{-1}}{2}\]
Now we know that $\sqrt{-1}=i$ therefore on substituting, we get:
\[\Rightarrow ({{x}_{1}},{{x}_{2}})=\dfrac{-8\pm 6i}{2}\]
Now on splitting the denominator, we get:
\[\Rightarrow ({{x}_{1}},{{x}_{2}})=-\dfrac{8}{2}\pm \dfrac{6i}{2}\]
Now on simplifying the terms, we get:
\[\Rightarrow ({{x}_{1}},{{x}_{2}})=-4\pm 3i\], which are the roots of the equation.
From the above expression we can see that ${{x}_{1}}=-4+3i$ and ${{x}_{2}}=-4-3i$ are the roots of the equation which is the required solution.

Note: A quadratic equation is a polynomial equation with a degree $2$, quadratic equations are used mostly in statistics when there is a power. It is not necessary that both the roots of the equation will be the same and some quadratic equation might not have proper roots. The quadratic equation can also be solved by splitting the middle term.