Solve sqrt{2r+3}=r

$$\sqrt{ 2r+3 } = r$$

Solve for r

$$r=3$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{2r+3}\right)^{2}=r^{2}$$

Calculate \sqrt{2r+3} to the power of 2 and get 2r+3.

$$2r+3=r^{2}$$

Subtract r^{2} from both sides.

$$2r+3-r^{2}=0$$

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

$$-r^{2}+2r+3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -r^{2}+ar+br+3. To find a and b, set up a system to be solved.

$$a+b=2 ab=-3=-3$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.

$$a=3 b=-1$$

Rewrite -r^{2}+2r+3 as \left(-r^{2}+3r\right)+\left(-r+3\right).

$$\left(-r^{2}+3r\right)+\left(-r+3\right)$$

Factor out -r in the first and -1 in the second group.

$$-r\left(r-3\right)-\left(r-3\right)$$

Factor out common term r-3 by using distributive property.

$$\left(r-3\right)\left(-r-1\right)$$

To find equation solutions, solve r-3=0 and -r-1=0.

$$r=3 r=-1$$

Substitute 3 for r in the equation \sqrt{2r+3}=r.

$$\sqrt{2\times 3+3}=3$$

Simplify. The value r=3 satisfies the equation.

$$3=3$$

Substitute -1 for r in the equation \sqrt{2r+3}=r.

$$\sqrt{2\left(-1\right)+3}=-1$$

Simplify. The value r=-1 does not satisfy the equation because the left and the right hand side have opposite signs.

$$1=-1$$

Equation \sqrt{2r+3}=r has a unique solution.

$$r=3$$