Solve 3+sqrt{2x-5}=x-1

$$3 + \sqrt { 2 x - 5 } = x - 1$$

Solve for x

$$x=7$$

Solution Steps

Subtract 3 from both sides of the equation.

$$\sqrt{2x-5}=x-1-3$$

Subtract 3 from -1 to get -4.

$$\sqrt{2x-5}=x-4$$

Square both sides of the equation.

$$\left(\sqrt{2x-5}\right)^{2}=\left(x-4\right)^{2}$$

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

$$2x-5=\left(x-4\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-4\right)^{2}.

$$2x-5=x^{2}-8x+16$$

Subtract x^{2} from both sides.

$$2x-5-x^{2}=-8x+16$$

Add 8x to both sides.

$$2x-5-x^{2}+8x=16$$

Combine 2x and 8x to get 10x.

$$10x-5-x^{2}=16$$

Subtract 16 from both sides.

$$10x-5-x^{2}-16=0$$

Subtract 16 from -5 to get -21.

$$10x-21-x^{2}=0$$

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

$$-x^{2}+10x-21=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-21. To find a and b, set up a system to be solved.

$$a+b=10 ab=-\left(-21\right)=21$$

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 21.

$$1,21 3,7$$

Calculate the sum for each pair.

$$1+21=22 3+7=10$$

The solution is the pair that gives sum 10.

$$a=7 b=3$$

Rewrite -x^{2}+10x-21 as \left(-x^{2}+7x\right)+\left(3x-21\right).

$$\left(-x^{2}+7x\right)+\left(3x-21\right)$$

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

$$-x\left(x-7\right)+3\left(x-7\right)$$

Factor out common term x-7 by using distributive property.

$$\left(x-7\right)\left(-x+3\right)$$

To find equation solutions, solve x-7=0 and -x+3=0.

$$x=7 x=3$$

Substitute 7 for x in the equation 3+\sqrt{2x-5}=x-1.

$$3+\sqrt{2\times 7-5}=7-1$$

Simplify. The value x=7 satisfies the equation.

$$6=6$$

Substitute 3 for x in the equation 3+\sqrt{2x-5}=x-1.

$$3+\sqrt{2\times 3-5}=3-1$$

Simplify. The value x=3 does not satisfy the equation.

$$4=2$$

Equation \sqrt{2x-5}=x-4 has a unique solution.

$$x=7$$