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

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

Solve for x

$$x=13$$

Solution Steps

Square both sides of the equation.

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

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

$$x^{2}-10x+25=\left(2\sqrt{x+3}\right)^{2}$$

Expand \left(2\sqrt{x+3}\right)^{2}.

$$x^{2}-10x+25=2^{2}\left(\sqrt{x+3}\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$x^{2}-10x+25=4\left(\sqrt{x+3}\right)^{2}$$

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

$$x^{2}-10x+25=4\left(x+3\right)$$

Use the distributive property to multiply 4 by x+3.

$$x^{2}-10x+25=4x+12$$

Subtract 4x from both sides.

$$x^{2}-10x+25-4x=12$$

Combine -10x and -4x to get -14x.

$$x^{2}-14x+25=12$$

Subtract 12 from both sides.

$$x^{2}-14x+25-12=0$$

Subtract 12 from 25 to get 13.

$$x^{2}-14x+13=0$$

To solve the equation, factor x^{2}-14x+13 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.

$$a+b=-14 ab=13$$

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.

$$a=-13 b=-1$$

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

$$\left(x-13\right)\left(x-1\right)$$

To find equation solutions, solve x-13=0 and x-1=0.

$$x=13 x=1$$

Substitute 13 for x in the equation x-5=2\sqrt{x+3}.

$$13-5=2\sqrt{13+3}$$

Simplify. The value x=13 satisfies the equation.

$$8=8$$

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

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

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

$$-4=4$$

Equation x-5=2\sqrt{x+3} has a unique solution.

$$x=13$$