Question

How do you solve $ {{x}^{2}}+16=0 $ using the quadratic formula?

Answer 1

Hint: In this question, we are given a quadratic equation and we need to find the value of x using quadratic formula. For this we will first compare $ {{x}^{2}}+16 $ with $ a{{x}^{2}}+bx+c $ to find the value of a, b, c. Then we will use the quadratic formula $ x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} $ to find the value of x. We will also use the form that $ \sqrt{-1}=i\text{ and }\sqrt{x\cdot y}=\sqrt{x}\cdot \sqrt{y} $ .

Complete step by step answer:
Here we are given the quadratic equation (equation of degree 2) as $ {{x}^{2}}+16=0 $ . We need to solve it to find the value of x using the quadratic formula since the equation is of degree 2, so we will obtain two values of x satisfying this equation.
Let us first understand the quadratic formula, for an equation of the degree two of the form $ a{{x}^{2}}+bx+c=0 $ value of x are given by the formula called quadratic formula which is equal to $ x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a} $ .
Now we are given the equation as $ {{x}^{2}}+16=0 $ comparing it with general equation $ a{{x}^{2}}+bx+c=0 $ we get a = 1, b = 0 and c = 16.
We have obtained the value of a, b, c. So let us put them in the quadratic formula. We get, $ x=\dfrac{-0\pm \sqrt{{{\left( 0 \right)}^{2}}-4\left( 1 \right)\left( 16 \right)}}{2\left( 1 \right)}\Rightarrow x=\dfrac{\pm \sqrt{-64}}{2} $ .
By the law of radicals, we can write $ \sqrt{xy} $ as $ \sqrt{x}\sqrt{y} $ . So applying it on $ \sqrt{-64} $ we get, $ x=\dfrac{\pm \sqrt{-1.64}}{2}\Rightarrow x=\dfrac{\pm \sqrt{-1}.\sqrt{64}}{2} $ .
Now we know that, $ 8\times 8=64\Rightarrow \sqrt{64}=8 $ so putting in this value we get, $ x=\dfrac{\pm \sqrt{-1}.8}{2} $ .
We know $ \sqrt{-1} $ is an imaginary number, and is denoted by i. Therefore, $ \sqrt{-1}=i $ so our value becomes $ x=\dfrac{\pm i.8}{2} $ .
Simplifying we get $ x=\pm 4i $ .
The two values of x become 4i and -4i.
Therefore the required value of x is 4i and -4i.

Note:
Students should take care of signs while comparing the equation. They often get confused between variable b and c. They should not ignore negative signs inside the root. Students can check their answers by putting the obtained values back into the equation to see if they satisfy the equation.