Chapter 12.3, Problem 14WE

To determine

To solve: the equation $j^2-6j=13$ using quadratic formula. Give irrational roots in simplest radical form and then approximate them to the nearest tenth.

Answer to Problem 14WE

The required answer is: solution set is 7.69, - 1.69

Explanation of Solution

Given information:

The given equation: $j^2-6j=13$

Calculation:

  $\begin{array}{l}{j}^2-6j=13\\ j^2-6j-13=13-13\end{array}$

(Subtract 13 from both sides)

  $j^2-6j-13=0$

(Simplify)

  $j=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

(Quadratic formula)

Where $a=1,b=-6andc=-13$

  $j=\frac{-(-6)\pm \sqrt{{(-6)}^2-4(1)(-13)}}{2(1)}$

(Remove the brackets)

  $\begin{array}{l}=\frac{6\pm \sqrt{36+52}}{2}\\ =\frac{6\pm \sqrt{88}}{2}\\ =\frac{6\pm \sqrt{4\cdot 22}}{2}\end{array}$

(Product property of square roots)

  $\begin{array}{l}=\frac{6\pm 2\sqrt{22}}{2}\\ =\frac{2(3\pm \sqrt{22})}{2}\end{array}$

(Distributive property)

  $=3\pm \sqrt{22}$

(Simplify)

Since $\sqrt{22}\approx 4.69$ , $j\approx 3+4.69\approx 7.69$

Or   $j\approx 3-4.69\approx -1.69$

Therefore the solution set is $3+\sqrt{22},3-\sqrt{22}$ or 7.69, - 1.69