Solve (x-7)^2=0

$$( x - 7 ) ^ { 2 } = 0$$

Solve for x

$$x=7$$

Steps Using Factoring

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-7\right)^{2}.

$$x^{2}-14x+49=0$$

To solve the equation, factor x^{2}-14x+49 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.

$$a+b=-14 ab=49$$

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 49.

$$-1,-49 -7,-7$$

Calculate the sum for each pair.

$$-1-49=-50 -7-7=-14$$

The solution is the pair that gives sum -14.

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

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

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

Rewrite as a binomial square.

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

To find equation solution, solve x-7=0.

$$x=7$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-7\right)^{2}.

$$x^{2}-14x+49=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+49. To find a and b, set up a system to be solved.

$$a+b=-14 ab=1\times 49=49$$

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 49.

$$-1,-49 -7,-7$$

Calculate the sum for each pair.

$$-1-49=-50 -7-7=-14$$

The solution is the pair that gives sum -14.

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

Rewrite x^{2}-14x+49 as \left(x^{2}-7x\right)+\left(-7x+49\right).

$$\left(x^{2}-7x\right)+\left(-7x+49\right)$$

Factor out x in the first and -7 in the second group.

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

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

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

Rewrite as a binomial square.

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

To find equation solution, solve x-7=0.

$$x=7$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-7\right)^{2}.

$$x^{2}-14x+49=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -14 for b, and 49 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-14\right)±\sqrt{\left(-14\right)^{2}-4\times 49}}{2}$$

Square -14.

$$x=\frac{-\left(-14\right)±\sqrt{196-4\times 49}}{2}$$

Multiply -4 times 49.

$$x=\frac{-\left(-14\right)±\sqrt{196-196}}{2}$$

Add 196 to -196.

$$x=\frac{-\left(-14\right)±\sqrt{0}}{2}$$

Take the square root of 0.

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

The opposite of -14 is 14.

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

Divide 14 by 2.

$$x=7$$

Steps for Completing the Square

Take the square root of both sides of the equation.

$$\sqrt{\left(x-7\right)^{2}}=\sqrt{0}$$

Simplify.

$$x-7=0 x-7=0$$

Add 7 to both sides of the equation.

$$x=7 x=7$$

The equation is now solved. Solutions are the same.

$$x=7$$