Solve sqrt{2x+5}=x-5

$$\sqrt { 2 x + 5 } = x - 5$$

Solve for x

$$x=10$$

Solution Steps

Square both sides of the equation.

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

Calculate \sqrt{2x+5} to the power of 2 and get 2x+5.

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

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

$$2x+5=x^{2}-10x+25$$

Subtract x^{2} from both sides.

$$2x+5-x^{2}=-10x+25$$

Add 10x to both sides.

$$2x+5-x^{2}+10x=25$$

Combine 2x and 10x to get 12x.

$$12x+5-x^{2}=25$$

Subtract 25 from both sides.

$$12x+5-x^{2}-25=0$$

Subtract 25 from 5 to get -20.

$$12x-20-x^{2}=0$$

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

$$-x^{2}+12x-20=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-20. To find a and b, set up a system to be solved.

$$a+b=12 ab=-\left(-20\right)=20$$

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 20.

$$1,20 2,10 4,5$$

Calculate the sum for each pair.

$$1+20=21 2+10=12 4+5=9$$

The solution is the pair that gives sum 12.

$$a=10 b=2$$

Rewrite -x^{2}+12x-20 as \left(-x^{2}+10x\right)+\left(2x-20\right).

$$\left(-x^{2}+10x\right)+\left(2x-20\right)$$

Factor out -x in the first and 2 in the second group.

$$-x\left(x-10\right)+2\left(x-10\right)$$

Factor out common term x-10 by using distributive property.

$$\left(x-10\right)\left(-x+2\right)$$

To find equation solutions, solve x-10=0 and -x+2=0.

$$x=10 x=2$$

Substitute 10 for x in the equation \sqrt{2x+5}=x-5.

$$\sqrt{2\times 10+5}=10-5$$

Simplify. The value x=10 satisfies the equation.

$$5=5$$

Substitute 2 for x in the equation \sqrt{2x+5}=x-5.

$$\sqrt{2\times 2+5}=2-5$$

Simplify. The value x=2 does not satisfy the equation because the left and the right hand side have opposite signs.

$$3=-3$$

Equation \sqrt{2x+5}=x-5 has a unique solution.

$$x=10$$