Question

solve $x^2 - 8x = 20$ by completing the square. which is the solution set of the equation?
$\bigcirc\\ \\{-22, 14\\}$
$\bigcirc\\ \\{-14, 22\\}$
$\bigcirc\\ \\{-10, 2\\}$
$\bigcirc\\ \\{-2, 10\\}$

Explanation:

Step1: Complete the square for \(x^2 - 8x\)

To complete the square for the quadratic expression \(x^2 - 8x\), we take half of the coefficient of \(x\), which is \(\frac{-8}{2}=-4\), and then square it: \((-4)^2 = 16\). We add and subtract this value (or just add it to both sides of the equation to maintain equality). So, add 16 to both sides of the equation \(x^2 - 8x = 20\):

$$x^2 - 8x + 16 = 20 + 16$$

Step2: Rewrite the left side as a square

The left side \(x^2 - 8x + 16\) is a perfect square trinomial and can be written as \((x - 4)^2\). The right side simplifies to \(36\):

$$(x - 4)^2 = 36$$

Step3: Take the square root of both sides

Take the square root of both sides. Remember that when we take the square root of a number, we consider both the positive and negative roots:

$$x - 4=\pm\sqrt{36}$$

Since \(\sqrt{36} = 6\), we have:

$$x - 4=\pm6$$

Step4: Solve for \(x\)

Solve for \(x\) by adding 4 to both sides of each equation.
For the positive root: \(x=4 + 6=10\)
For the negative root: \(x=4-6=-2\)

Answer:

\(\{-2, 10\}\)