Question

How do you solve $ {x^2} + 8 = 0 $ using quadratic formula?

Answer 1

Hint: A polynomial of degree two is called a quadratic polynomial and its zeros can be found using many methods like factorization, completing the square, graphs, quadratic formula etc. The quadratic formula is used when we fail to find the factors of the equation. In the given question we have to solve the given quadratic equation using the quadratic formula, so for that we will have to first bring the given equation in the standard equation form and then plug in the values of the coefficients in the quadratic formula.

Complete step-by-step answer:
The equation given is $ {x^2} + 8 = 0 $
On comparing the given equation with the standard quadratic equation $ a{x^2} + bx + c = 0 $ , we get –
 $ a = 1,\,b = 0,\,c = 8 $
The Quadratic formula is given as –
 $ x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} $
Putting the values in the above equation, we get –
 $
  x = \dfrac{{ - 0 \pm \sqrt {{0^2} - 4(1)( 8)} }}{{2(1)}} \\
   \Rightarrow x = \pm \dfrac{{\sqrt {-32} }}{2} \\
   \Rightarrow x = \pm \dfrac{{i4\sqrt 2 }}{2} \\
   \Rightarrow x = \pm i2\sqrt 2 \;
  $
Hence the zeros of the given equation are $ 2\sqrt 2 $ and $ - 2\sqrt 2 $ .
So, the correct answer is “ $ i2\sqrt 2 $ and $ - i2\sqrt 2 $ ”.

Note: A polynomial equation is an algebraic expression that contains numerical values and alphabets too, but when the alphabets representing an unknown variable quantity are raised to some power such that the exponent is a non-negative integer, the algebraic expression becomes a polynomial equation. The highest exponent of the polynomial in a polynomial equation is called its degree. Now, various fields of mathematics require the point at which the value of a polynomial is zero, those values are called the factors/solution/zeros of the given polynomial. On the x-axis, the value of y is zero so the roots of an equation are the points on the x-axis, that is the roots are simply the x-intercepts.