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

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

Solve for x

$$x=5$$

Solution Steps

Subtract 5 from both sides of the equation.

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

Subtract 5 from 3 to get -2.

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

Square both sides of the equation.

$$\left(x-2\right)^{2}=\left(\sqrt{2x-1}\right)^{2}$$

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

$$x^{2}-4x+4=\left(\sqrt{2x-1}\right)^{2}$$

Calculate \sqrt{2x-1} to the power of 2 and get 2x-1.

$$x^{2}-4x+4=2x-1$$

Subtract 2x from both sides.

$$x^{2}-4x+4-2x=-1$$

Combine -4x and -2x to get -6x.

$$x^{2}-6x+4=-1$$

Add 1 to both sides.

$$x^{2}-6x+4+1=0$$

Add 4 and 1 to get 5.

$$x^{2}-6x+5=0$$

To solve the equation, factor x^{2}-6x+5 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=-6 ab=5$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. The only such pair is the system solution.

$$a=-5 b=-1$$

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

$$\left(x-5\right)\left(x-1\right)$$

To find equation solutions, solve x-5=0 and x-1=0.

$$x=5 x=1$$

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

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

Simplify. The value x=5 satisfies the equation.

$$8=8$$

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

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

Simplify. The value x=1 does not satisfy the equation.

$$4=6$$

Equation x-2=\sqrt{2x-1} has a unique solution.

$$x=5$$