Solve r-1=sqrt{8r+120}

$$r - 1 = \sqrt { 8 r + 120 }$$

Solve for r

$$r=17$$

Solution Steps

Square both sides of the equation.

$$\left(r-1\right)^{2}=\left(\sqrt{8r+120}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(r-1\right)^{2}.

$$r^{2}-2r+1=\left(\sqrt{8r+120}\right)^{2}$$

Calculate \sqrt{8r+120} to the power of 2 and get 8r+120.

$$r^{2}-2r+1=8r+120$$

Subtract 8r from both sides.

$$r^{2}-2r+1-8r=120$$

Combine -2r and -8r to get -10r.

$$r^{2}-10r+1=120$$

Subtract 120 from both sides.

$$r^{2}-10r+1-120=0$$

Subtract 120 from 1 to get -119.

$$r^{2}-10r-119=0$$

To solve the equation, factor r^{2}-10r-119 using formula r^{2}+\left(a+b\right)r+ab=\left(r+a\right)\left(r+b\right). To find a and b, set up a system to be solved.

$$a+b=-10 ab=-119$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -119.

$$1,-119 7,-17$$

Calculate the sum for each pair.

$$1-119=-118 7-17=-10$$

The solution is the pair that gives sum -10.

$$a=-17 b=7$$

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

$$\left(r-17\right)\left(r+7\right)$$

To find equation solutions, solve r-17=0 and r+7=0.

$$r=17 r=-7$$

Substitute 17 for r in the equation r-1=\sqrt{8r+120}.

$$17-1=\sqrt{8\times 17+120}$$

Simplify. The value r=17 satisfies the equation.

$$16=16$$

Substitute -7 for r in the equation r-1=\sqrt{8r+120}.

$$-7-1=\sqrt{8\left(-7\right)+120}$$

Simplify. The value r=-7 does not satisfy the equation because the left and the right hand side have opposite signs.

$$-8=8$$

Equation r-1=\sqrt{8r+120} has a unique solution.

$$r=17$$