Solve sqrt{9+8x}+sqrt{6x+7}=2

$$\sqrt { 9 + 8 x } + \sqrt { 6 x + 7 } = 2$$

Solve for x

$$x=-1$$

Solution Steps

Subtract \sqrt{6x+7} from both sides of the equation.

$$\sqrt{9+8x}=2-\sqrt{6x+7}$$

Square both sides of the equation.

$$\left(\sqrt{9+8x}\right)^{2}=\left(2-\sqrt{6x+7}\right)^{2}$$

Calculate \sqrt{9+8x} to the power of 2 and get 9+8x.

$$9+8x=\left(2-\sqrt{6x+7}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2-\sqrt{6x+7}\right)^{2}.

$$9+8x=4-4\sqrt{6x+7}+\left(\sqrt{6x+7}\right)^{2}$$

Calculate \sqrt{6x+7} to the power of 2 and get 6x+7.

$$9+8x=4-4\sqrt{6x+7}+6x+7$$

Add 4 and 7 to get 11.

$$9+8x=11-4\sqrt{6x+7}+6x$$

Subtract 11+6x from both sides of the equation.

$$9+8x-\left(11+6x\right)=-4\sqrt{6x+7}$$

To find the opposite of 11+6x, find the opposite of each term.

$$9+8x-11-6x=-4\sqrt{6x+7}$$

Subtract 11 from 9 to get -2.

$$-2+8x-6x=-4\sqrt{6x+7}$$

Combine 8x and -6x to get 2x.

$$-2+2x=-4\sqrt{6x+7}$$

Square both sides of the equation.

$$\left(-2+2x\right)^{2}=\left(-4\sqrt{6x+7}\right)^{2}$$

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

$$4-8x+4x^{2}=\left(-4\sqrt{6x+7}\right)^{2}$$

Expand \left(-4\sqrt{6x+7}\right)^{2}.

$$4-8x+4x^{2}=\left(-4\right)^{2}\left(\sqrt{6x+7}\right)^{2}$$

Calculate -4 to the power of 2 and get 16.

$$4-8x+4x^{2}=16\left(\sqrt{6x+7}\right)^{2}$$

Calculate \sqrt{6x+7} to the power of 2 and get 6x+7.

$$4-8x+4x^{2}=16\left(6x+7\right)$$

Use the distributive property to multiply 16 by 6x+7.

$$4-8x+4x^{2}=96x+112$$

Subtract 96x from both sides.

$$4-8x+4x^{2}-96x=112$$

Combine -8x and -96x to get -104x.

$$4-104x+4x^{2}=112$$

Subtract 112 from both sides.

$$4-104x+4x^{2}-112=0$$

Subtract 112 from 4 to get -108.

$$-108-104x+4x^{2}=0$$

Divide both sides by 4.

$$-27-26x+x^{2}=0$$

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

$$x^{2}-26x-27=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-27. To find a and b, set up a system to be solved.

$$a+b=-26 ab=1\left(-27\right)=-27$$

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. List all such integer pairs that give product -27.

$$1,-27 3,-9$$

Calculate the sum for each pair.

$$1-27=-26 3-9=-6$$

The solution is the pair that gives sum -26.

$$a=-27 b=1$$

Rewrite x^{2}-26x-27 as \left(x^{2}-27x\right)+\left(x-27\right).

$$\left(x^{2}-27x\right)+\left(x-27\right)$$

Factor out x in x^{2}-27x.

$$x\left(x-27\right)+x-27$$

Factor out common term x-27 by using distributive property.

$$\left(x-27\right)\left(x+1\right)$$

To find equation solutions, solve x-27=0 and x+1=0.

$$x=27 x=-1$$

Substitute 27 for x in the equation \sqrt{9+8x}+\sqrt{6x+7}=2.

$$\sqrt{9+8\times 27}+\sqrt{6\times 27+7}=2$$

Simplify. The value x=27 does not satisfy the equation.

$$28=2$$

Substitute -1 for x in the equation \sqrt{9+8x}+\sqrt{6x+7}=2.

$$\sqrt{9+8\left(-1\right)}+\sqrt{6\left(-1\right)+7}=2$$

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

$$2=2$$

Equation \sqrt{8x+9}=-\sqrt{6x+7}+2 has a unique solution.

$$x=-1$$