Solve x+5=sqrt{12x+28}

$$x + 5 = \sqrt { 12 x + 28 }$$

Solve for x

$$x=3$$

$$x=-1$$

Solution Steps

Square both sides of the equation.

$$\left(x+5\right)^{2}=\left(\sqrt{12x+28}\right)^{2}$$

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

$$x^{2}+10x+25=\left(\sqrt{12x+28}\right)^{2}$$

Calculate \sqrt{12x+28} to the power of 2 and get 12x+28.

$$x^{2}+10x+25=12x+28$$

Subtract 12x from both sides.

$$x^{2}+10x+25-12x=28$$

Combine 10x and -12x to get -2x.

$$x^{2}-2x+25=28$$

Subtract 28 from both sides.

$$x^{2}-2x+25-28=0$$

Subtract 28 from 25 to get -3.

$$x^{2}-2x-3=0$$

To solve the equation, factor x^{2}-2x-3 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.

$$a+b=-2 ab=-3$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. The only such pair is the system solution.

$$a=-3 b=1$$

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

$$\left(x-3\right)\left(x+1\right)$$

To find equation solutions, solve x-3=0 and x+1=0.

$$x=3 x=-1$$

Substitute 3 for x in the equation x+5=\sqrt{12x+28}.

$$3+5=\sqrt{12\times 3+28}$$

Simplify. The value x=3 satisfies the equation.

$$8=8$$

Substitute -1 for x in the equation x+5=\sqrt{12x+28}.

$$-1+5=\sqrt{12\left(-1\right)+28}$$

Simplify. The value x=-1 satisfies the equation.

$$4=4$$

List all solutions of x+5=\sqrt{12x+28}.

$$x=3 x=-1$$