Solve (1)(x-5)^2=4

$$( 1 ) ( x - 5 ) ^ { 2 } = 4$$

Solve for x

$$x=7$$

$$x=3$$

Steps Using Factoring

Divide both sides by 1.

$$\left(x-5\right)^{2}=4$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-5\right)^{2}.

$$x^{2}-10x+25=4$$

Subtract 4 from both sides.

$$x^{2}-10x+25-4=0$$

Subtract 4 from 25 to get 21.

$$x^{2}-10x+21=0$$

To solve the equation, factor x^{2}-10x+21 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=-10 ab=21$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 21.

$$-1,-21 -3,-7$$

Calculate the sum for each pair.

$$-1-21=-22 -3-7=-10$$

The solution is the pair that gives sum -10.

$$a=-7 b=-3$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-7\right)\left(x-3\right)$$

To find equation solutions, solve x-7=0 and x-3=0.

$$x=7 x=3$$

Steps Using Factoring By Grouping

Divide both sides by 1.

$$\left(x-5\right)^{2}=4$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-5\right)^{2}.

$$x^{2}-10x+25=4$$

Subtract 4 from both sides.

$$x^{2}-10x+25-4=0$$

Subtract 4 from 25 to get 21.

$$x^{2}-10x+21=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+21. To find a and b, set up a system to be solved.

$$a+b=-10 ab=1\times 21=21$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 21.

$$-1,-21 -3,-7$$

Calculate the sum for each pair.

$$-1-21=-22 -3-7=-10$$

The solution is the pair that gives sum -10.

$$a=-7 b=-3$$

Rewrite x^{2}-10x+21 as \left(x^{2}-7x\right)+\left(-3x+21\right).

$$\left(x^{2}-7x\right)+\left(-3x+21\right)$$

Factor out x in the first and -3 in the second group.

$$x\left(x-7\right)-3\left(x-7\right)$$

Factor out common term x-7 by using distributive property.

$$\left(x-7\right)\left(x-3\right)$$

To find equation solutions, solve x-7=0 and x-3=0.

$$x=7 x=3$$

Steps Using the Quadratic Formula

Divide both sides by 1.

$$\left(x-5\right)^{2}=4$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-5\right)^{2}.

$$x^{2}-10x+25=4$$

Subtract 4 from both sides.

$$x^{2}-10x+25-4=0$$

Subtract 4 from 25 to get 21.

$$x^{2}-10x+21=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -10 for b, and 21 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 21}}{2}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\times 21}}{2}$$

Multiply -4 times 21.

$$x=\frac{-\left(-10\right)±\sqrt{100-84}}{2}$$

Add 100 to -84.

$$x=\frac{-\left(-10\right)±\sqrt{16}}{2}$$

Take the square root of 16.

$$x=\frac{-\left(-10\right)±4}{2}$$

The opposite of -10 is 10.

$$x=\frac{10±4}{2}$$

Now solve the equation x=\frac{10±4}{2} when ± is plus. Add 10 to 4.

$$x=\frac{14}{2}$$

Divide 14 by 2.

$$x=7$$

Now solve the equation x=\frac{10±4}{2} when ± is minus. Subtract 4 from 10.

$$x=\frac{6}{2}$$

Divide 6 by 2.

$$x=3$$

The equation is now solved.

$$x=7 x=3$$

Steps for Completing the Square

Divide both sides by 1.

$$\left(x-5\right)^{2}=4$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-5\right)^{2}}=\sqrt{4}$$

Simplify.

$$x-5=2 x-5=-2$$

Add 5 to both sides of the equation.

$$x=7 x=3$$