Question

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

Answer 1

Hint:Here, we will compare the given quadratic equation with the general formula of the quadratic equation. Then, comparing the coefficients and substituting them in the quadratic formula, we will be able to solve the given equation further. A quadratic equation is an equation that has the highest degree of 2 and has two solutions. As the given equation is a quadratic equation, we will get two solutions, which can be either positive or negative.

Formula Used:
\[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\], where, \[a,b,c\]are the coefficients of \[{x^2},x\] and the constant respectively.

Complete step-by-step answer:
Given quadratic equation is \[{x^2} + 16 = 0\]
Now, in order to solve this equation with the help of quadratic formula, we will compare this equation with the general quadratic equation i.e. \[a{x^2} + bx + c = 0\]
Hence, we have,
\[a = 1\], \[b = 0\] and \[c = 16\]
Now, substituting \[a = 1\], \[b = 0\] and \[c = 16\] in the formula \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\], we get,
\[x = \dfrac{{ - 0 \pm \sqrt {{0^2} - 4\left( 1 \right)\left( {16} \right)} }}{{2\left( 1 \right)}}\]
Simplifying the expression, we get
\[ \Rightarrow x = \dfrac{{ \pm 8\sqrt { - 1} }}{2}\]
Here, substituting \[\sqrt { - 1} = i\] as it is imaginary, and dividing the numerator and denominator by 2, we get,
\[ \Rightarrow x = \pm 4i\]
Thus, this is the required answer.

Note: If there is any negative number inside the square root we denote it as iota and it is known as a complex number. If in the question, it was not mentioned that we have to use the quadratic formula, then, we could have used the alternate way of solving this equation.
Given quadratic equation is:
\[{x^2} + 16 = 0\]
Subtracting 16 from both the sides, we get,
\[{x^2} = - 16\]
Taking square root on both sides,
\[ \Rightarrow x = \pm \sqrt { - 16} \]
\[ \Rightarrow x = \pm \sqrt { - 1 \times 4 \times 4} \]
Taking out 4 from the square root, we get,
\[ \Rightarrow x = \pm 4\sqrt { - 1} \]
Here, substituting \[\sqrt { - 1} = i\] as it is a complex number or an imaginary number, we get,
\[ \Rightarrow x = \pm 4i\]
Hence, this is the required answer.