Chapter 5.7, Problem 20PPS

To determine

Solve the given equation. State the number and type of roots.

Answer to Problem 20PPS

The equation $x^3+1=0$ has three roots, one is real 1, other two are imaginary.

Explanation of Solution

Given information:

The equation has two roots, both are imaginary $\frac{25+\sqrt{-87}}{4}$ and $\frac{25-\sqrt{-87}}{4}$ .

Formula used:

• For equation $a{x}^2+bx+c=0$,

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Discriminant: $\sqrt{b^2-4ac}$

If $\sqrt{b^2-4ac}>0$ , then given equation has real roots.

If $\sqrt{b^2-4ac}=0$ , then given equation has real and equal roots.

If $\sqrt{b^2-4ac}<0$ , then given equation has imaginary roots.

  $2x^2-5x+14=0$

Finding discriminant for above equation,

  $\begin{array}{l}\sqrt{b^2-4ac}\\ \\ =\sqrt{{\left(-5\right)}^2-4\cdot 14\cdot 2}\\ \\ =\sqrt{25-112}\\ \\ =\sqrt{-87}\end{array}$

As the discriminant $\sqrt{b^2-4ac}<0$ , given equation has imaginary roots.

Recall, For equation $a{x}^2+bx+c=0$ , $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ ,

  $\begin{array}{l}x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ \\ x=\frac{-\left(-25\right)\pm \sqrt{-87}}{2\cdot 2}\\ \\ x=\frac{25\pm \sqrt{-87}}{4}\end{array}$

The equation has two roots, both are imaginary $\frac{25+\sqrt{-87}}{4}$ and $\frac{25-\sqrt{-87}}{4}$ .