Solve sqrt{y-7}=y-7

$$\sqrt { y - 7 } = y - 7$$

Solve for y

$$y=7$$

$$y=8$$

Solution Steps

Square both sides of the equation.

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

Calculate \sqrt{y-7} to the power of 2 and get y-7.

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

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

$$y-7=y^{2}-14y+49$$

Subtract y^{2} from both sides.

$$y-7-y^{2}=-14y+49$$

Add 14y to both sides.

$$y-7-y^{2}+14y=49$$

Combine y and 14y to get 15y.

$$15y-7-y^{2}=49$$

Subtract 49 from both sides.

$$15y-7-y^{2}-49=0$$

Subtract 49 from -7 to get -56.

$$15y-56-y^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-y^{2}+15y-56=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -y^{2}+ay+by-56. To find a and b, set up a system to be solved.

$$a+b=15 ab=-\left(-56\right)=56$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 56.

$$1,56 2,28 4,14 7,8$$

Calculate the sum for each pair.

$$1+56=57 2+28=30 4+14=18 7+8=15$$

The solution is the pair that gives sum 15.

$$a=8 b=7$$

Rewrite -y^{2}+15y-56 as \left(-y^{2}+8y\right)+\left(7y-56\right).

$$\left(-y^{2}+8y\right)+\left(7y-56\right)$$

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

$$-y\left(y-8\right)+7\left(y-8\right)$$

Factor out common term y-8 by using distributive property.

$$\left(y-8\right)\left(-y+7\right)$$

To find equation solutions, solve y-8=0 and -y+7=0.

$$y=8 y=7$$

Substitute 8 for y in the equation \sqrt{y-7}=y-7.

$$\sqrt{8-7}=8-7$$

Simplify. The value y=8 satisfies the equation.

$$1=1$$

Substitute 7 for y in the equation \sqrt{y-7}=y-7.

$$\sqrt{7-7}=7-7$$

Simplify. The value y=7 satisfies the equation.

$$0=0$$

List all solutions of \sqrt{y-7}=y-7.

$$y=8 y=7$$