Solve sqrt{3x+12}-1=sqrt{5x+9}

$$\sqrt{ 3x+12 } -1= \sqrt{ 5x+9 }$$

Solve for x

$$x=-1$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{3x+12}-1\right)^{2}=\left(\sqrt{5x+9}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{3x+12}-1\right)^{2}.

$$\left(\sqrt{3x+12}\right)^{2}-2\sqrt{3x+12}+1=\left(\sqrt{5x+9}\right)^{2}$$

Calculate \sqrt{3x+12} to the power of 2 and get 3x+12.

$$3x+12-2\sqrt{3x+12}+1=\left(\sqrt{5x+9}\right)^{2}$$

Add 12 and 1 to get 13.

$$3x+13-2\sqrt{3x+12}=\left(\sqrt{5x+9}\right)^{2}$$

Calculate \sqrt{5x+9} to the power of 2 and get 5x+9.

$$3x+13-2\sqrt{3x+12}=5x+9$$

Subtract 3x+13 from both sides of the equation.

$$-2\sqrt{3x+12}=5x+9-\left(3x+13\right)$$

To find the opposite of 3x+13, find the opposite of each term.

$$-2\sqrt{3x+12}=5x+9-3x-13$$

Combine 5x and -3x to get 2x.

$$-2\sqrt{3x+12}=2x+9-13$$

Subtract 13 from 9 to get -4.

$$-2\sqrt{3x+12}=2x-4$$

Square both sides of the equation.

$$\left(-2\sqrt{3x+12}\right)^{2}=\left(2x-4\right)^{2}$$

Expand \left(-2\sqrt{3x+12}\right)^{2}.

$$\left(-2\right)^{2}\left(\sqrt{3x+12}\right)^{2}=\left(2x-4\right)^{2}$$

Calculate -2 to the power of 2 and get 4.

$$4\left(\sqrt{3x+12}\right)^{2}=\left(2x-4\right)^{2}$$

Calculate \sqrt{3x+12} to the power of 2 and get 3x+12.

$$4\left(3x+12\right)=\left(2x-4\right)^{2}$$

Use the distributive property to multiply 4 by 3x+12.

$$12x+48=\left(2x-4\right)^{2}$$

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

$$12x+48=4x^{2}-16x+16$$

Subtract 4x^{2} from both sides.

$$12x+48-4x^{2}=-16x+16$$

Add 16x to both sides.

$$12x+48-4x^{2}+16x=16$$

Combine 12x and 16x to get 28x.

$$28x+48-4x^{2}=16$$

Subtract 16 from both sides.

$$28x+48-4x^{2}-16=0$$

Subtract 16 from 48 to get 32.

$$28x+32-4x^{2}=0$$

Divide both sides by 4.

$$7x+8-x^{2}=0$$

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

$$-x^{2}+7x+8=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+8. To find a and b, set up a system to be solved.

$$a+b=7 ab=-8=-8$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -8.

$$-1,8 -2,4$$

Calculate the sum for each pair.

$$-1+8=7 -2+4=2$$

The solution is the pair that gives sum 7.

$$a=8 b=-1$$

Rewrite -x^{2}+7x+8 as \left(-x^{2}+8x\right)+\left(-x+8\right).

$$\left(-x^{2}+8x\right)+\left(-x+8\right)$$

Factor out -x in the first and -1 in the second group.

$$-x\left(x-8\right)-\left(x-8\right)$$

Factor out common term x-8 by using distributive property.

$$\left(x-8\right)\left(-x-1\right)$$

To find equation solutions, solve x-8=0 and -x-1=0.

$$x=8 x=-1$$

Substitute 8 for x in the equation \sqrt{3x+12}-1=\sqrt{5x+9}.

$$\sqrt{3\times 8+12}-1=\sqrt{5\times 8+9}$$

Simplify. The value x=8 does not satisfy the equation.

$$5=7$$

Substitute -1 for x in the equation \sqrt{3x+12}-1=\sqrt{5x+9}.

$$\sqrt{3\left(-1\right)+12}-1=\sqrt{5\left(-1\right)+9}$$

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

$$2=2$$

Equation \sqrt{3x+12}-1=\sqrt{5x+9} has a unique solution.

$$x=-1$$