Chapter 1.5, Problem 81E

To determine

The number of real solutions with the help of discriminant.

Answer to Problem 81E

There is only one real root for the equation

Explanation of Solution

Given information:

  $x^2+2.20x+1.21=0$

Formula used: -

When  $b^2-4ac=0$  there is one real root.

When  $b^2-4ac>0$  there are two real roots.

When  $b^2-4ac<0$  there are two complex roots.

  $x^2+2.20x+1.21=0$

Now, finding discriminant to find number of real solutions:

  $a=1,b=2.20,c=1.21$

  $\begin{array}{l}a=1,b=2.20,c=1.21\\ \sqrt{b^2-4ac}=\sqrt{{\left(2.20\right)}^2-4\left(1\right)\left(1.21\right)}\\ \sqrt{b^2-4ac}=\sqrt{4.84-4.84}\\ \sqrt{b^2-4ac}=0\end{array}$

Hence, there is only one real root for equation $x^2+2.20x+1.21=0$ .