Chapter 9, Problem 2CT

To determine

To solve the equation by using any method

Answer to Problem 2CT

The solution is $x=3-\sqrt{19},x=3+\sqrt{19}$ .

Explanation of Solution

Given information:

  $x^2-6x=10$

Formula used:

Quadratic formula is:

x = $\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Where $a$ is coefficient of $x^2$ , b is coefficient of x , and c is constant

Calculation:

The given equation is $x^2-6x=10$

Rearranging to standard form,

  $x^2-6x-10$

Here, $a=1,b=-6,c=-10$

Using quadratic formula and substituting value,

  $x=$$\frac{-(-6)\pm \sqrt{{(-6)}^2-4(1)(-10)}}{2(1)}$

Solving,

  $x=\frac{6\pm \sqrt{36+40}}{2}$

So,

  $x=\frac{6\pm \sqrt{76}}{2}$

And,

  $x=\frac{6\pm \sqrt{(2)(2)(19)}}{2}$

Taking 2 out,

  $x=\frac{6\pm 2\sqrt{19}}{2}$

So solutions are,

  $x=\frac{2(3-\sqrt{19})}{2},x=\frac{2(3+\sqrt{19})}{2}$

Solving,

  $x=3-\sqrt{19},x=3+\sqrt{19}$

Hence, the solution is $x=3-\sqrt{19},x=3+\sqrt{19}$ .