Chapter 12.3, Problem 31WE

To determine

To write: a quadratic equation whose roots are $2\pm \sqrt{5}$

Answer to Problem 31WE

The required answer is: quadratic equation is $x^2-4x-1=0$

Explanation of Solution

Given information:

The given roots: $2\pm \sqrt{5}$

Calculation:

The roots of the quadratic equation are

  $2+\sqrt{5}and2-\sqrt{5}$

Sum of the roots $=(2+\sqrt{5})+(2-\sqrt{5})$

  $\begin{array}{l}=2+\sqrt{5}+2-\sqrt{5}\\ =2+2+\sqrt{5}-\sqrt{5}\\ =4\end{array}$

(Simplify)

The sum of the roots of a quadratic equation $=\frac{-b}{a}$

Hence, $\frac{-b}{a}=\frac{4}{1}$

  $\frac{-b}{a}\cdot -1=\frac{4}{1}\cdot -1$

(Multiply -1 to both sides)

  $\frac{b}{a}=\frac{-4}{1}$

(Simplify)

  $a=1andb=-4$

Product of the roots $=(2+\sqrt{5})\cdot (2-\sqrt{5})$

  $\begin{array}{l}$ =4-2\sqrt{5}+2\sqrt{5}-{(\sqrt{5})}^2$=4-5\\ =-1\end{array}$

(Simplify)

The product of the roots of a quadratic equation $=\frac{c}{a}$

Hence, $\frac{c}{a}=\frac{-1}{1}$

  $c=-1anda=1$

Therefore, the quadratic equation is $x^2-4x-1=0$