Chapter 7.3, Problem 10OE

To determine

To explain: the roots of the quadratic equation $a{x}^2+bx+c=0$ are unequal when $b^2-4ac>0$ .

Explanation of Solution

Consider the given equation,

  $a{x}^2+bx+c=0$

Solve the equation in quadratic formula.

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Two roots of the equation will be

  $\frac{-b+\sqrt{b^2-4ac}}{2a}$ and $\frac{-b-\sqrt{b^2-4ac}}{2a}$

Case I:

Consider $b^2-4ac<0$ .

Negative under square root will give imaginary value.

In that case roots will be complex conjugate.

Case II Consider $b^2-4ac=0$ .

The roots will become

  $\frac{-b}{2a}$ and $\frac{-b}{2a}$

In that case roots will be equal.

Case III Consider $b^2-4ac>0$ .

The roots will become

  $\frac{-b+\sqrt{b^2-4ac}}{2a}$ and $\frac{-b-\sqrt{b^2-4ac}}{2a}$

In that case roots will be unequal.