Chapter 5.7, Problem 24PPS

To determine

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

Answer to Problem 24PPS

The equation $x^3-6x^2+7x=0$ has three roots, all three are real $\frac{6+\sqrt{6}}{2}$ , $\frac{6-\sqrt{6}}{2}$ and 0.

Explanation of Solution

Given information:

The equation $x^3-6x^2+7x=0$ .

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.

  $x^3-6x^2+7x=0$

Taking $x$ common from LHS,

  $x\left(x^2-6x+7\right)=0$

In above equation, either one or two of $x^2-6x+7$ and $x$ is zero,

  $x^2-6x+7=0$

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

  $\begin{array}{l}x=\frac{-\left(-6\right)\pm \sqrt{{\left(-6\right)}^2-4\cdot 1\cdot 7}}{2\cdot 1}\\ \\ x=\frac{6\pm \sqrt{36-28}}{2}\\ \\ x=\frac{6\pm \sqrt{6}}{2}\end{array}$

The equation $x^3-6x^2+7x=0$ has three roots, all three are real $\frac{6+\sqrt{6}}{2}$ , $\frac{6-\sqrt{6}}{2}$ and 0.