Solve sqrt{3x-5}=x-5

$$\sqrt { 3 x - 5 } = x - 5$$

Solve for x

$$x=10$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{3x-5}\right)^{2}=\left(x-5\right)^{2}$$

Calculate \sqrt{3x-5} to the power of 2 and get 3x-5.

$$3x-5=\left(x-5\right)^{2}$$

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

$$3x-5=x^{2}-10x+25$$

Subtract x^{2} from both sides.

$$3x-5-x^{2}=-10x+25$$

Add 10x to both sides.

$$3x-5-x^{2}+10x=25$$

Combine 3x and 10x to get 13x.

$$13x-5-x^{2}=25$$

Subtract 25 from both sides.

$$13x-5-x^{2}-25=0$$

Subtract 25 from -5 to get -30.

$$13x-30-x^{2}=0$$

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

$$-x^{2}+13x-30=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-30. To find a and b, set up a system to be solved.

$$a+b=13 ab=-\left(-30\right)=30$$

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 30.

$$1,30 2,15 3,10 5,6$$

Calculate the sum for each pair.

$$1+30=31 2+15=17 3+10=13 5+6=11$$

The solution is the pair that gives sum 13.

$$a=10 b=3$$

Rewrite -x^{2}+13x-30 as \left(-x^{2}+10x\right)+\left(3x-30\right).

$$\left(-x^{2}+10x\right)+\left(3x-30\right)$$

Factor out -x in the first and 3 in the second group.

$$-x\left(x-10\right)+3\left(x-10\right)$$

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

$$\left(x-10\right)\left(-x+3\right)$$

To find equation solutions, solve x-10=0 and -x+3=0.

$$x=10 x=3$$

Substitute 10 for x in the equation \sqrt{3x-5}=x-5.

$$\sqrt{3\times 10-5}=10-5$$

Simplify. The value x=10 satisfies the equation.

$$5=5$$

Substitute 3 for x in the equation \sqrt{3x-5}=x-5.

$$\sqrt{3\times 3-5}=3-5$$

Simplify. The value x=3 does not satisfy the equation because the left and the right hand side have opposite signs.

$$2=-2$$

Equation \sqrt{3x-5}=x-5 has a unique solution.

$$x=10$$