Solve sqrt{x+1}-sqrt{9-x}=sqrt{2x-12}

$$\sqrt { x + 1 } - \sqrt { 9 - x } = \sqrt { 2 x - 12 }$$

Solve for x

$$x=8$$

$$x=7$$

Solution Steps

Square both sides of the equation.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{x+1}-\sqrt{9-x}\right)^{2}.

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

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

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

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

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

Add 1 and 9 to get 10.

$$x+10-2\sqrt{x+1}\sqrt{9-x}-x=\left(\sqrt{2x-12}\right)^{2}$$

Combine x and -x to get 0.

$$10-2\sqrt{x+1}\sqrt{9-x}=\left(\sqrt{2x-12}\right)^{2}$$

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

$$10-2\sqrt{x+1}\sqrt{9-x}=2x-12$$

Subtract 10 from both sides of the equation.

$$-2\sqrt{x+1}\sqrt{9-x}=2x-12-10$$

Subtract 10 from -12 to get -22.

$$-2\sqrt{x+1}\sqrt{9-x}=2x-22$$

Square both sides of the equation.

$$\left(-2\sqrt{x+1}\sqrt{9-x}\right)^{2}=\left(2x-22\right)^{2}$$

Expand \left(-2\sqrt{x+1}\sqrt{9-x}\right)^{2}.

$$\left(-2\right)^{2}\left(\sqrt{x+1}\right)^{2}\left(\sqrt{9-x}\right)^{2}=\left(2x-22\right)^{2}$$

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

$$4\left(\sqrt{x+1}\right)^{2}\left(\sqrt{9-x}\right)^{2}=\left(2x-22\right)^{2}$$

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

$$4\left(x+1\right)\left(\sqrt{9-x}\right)^{2}=\left(2x-22\right)^{2}$$

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

$$4\left(x+1\right)\left(9-x\right)=\left(2x-22\right)^{2}$$

Use the distributive property to multiply 4 by x+1.

$$\left(4x+4\right)\left(9-x\right)=\left(2x-22\right)^{2}$$

Apply the distributive property by multiplying each term of 4x+4 by each term of 9-x.

$$36x-4x^{2}+36-4x=\left(2x-22\right)^{2}$$

Combine 36x and -4x to get 32x.

$$32x-4x^{2}+36=\left(2x-22\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-22\right)^{2}.

$$32x-4x^{2}+36=4x^{2}-88x+484$$

Subtract 4x^{2} from both sides.

$$32x-4x^{2}+36-4x^{2}=-88x+484$$

Combine -4x^{2} and -4x^{2} to get -8x^{2}.

$$32x-8x^{2}+36=-88x+484$$

Add 88x to both sides.

$$32x-8x^{2}+36+88x=484$$

Combine 32x and 88x to get 120x.

$$120x-8x^{2}+36=484$$

Subtract 484 from both sides.

$$120x-8x^{2}+36-484=0$$

Subtract 484 from 36 to get -448.

$$120x-8x^{2}-448=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-8x^{2}+120x-448=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -8 for a, 120 for b, and -448 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-120±\sqrt{120^{2}-4\left(-8\right)\left(-448\right)}}{2\left(-8\right)}$$

Square 120.

$$x=\frac{-120±\sqrt{14400-4\left(-8\right)\left(-448\right)}}{2\left(-8\right)}$$

Multiply -4 times -8.

$$x=\frac{-120±\sqrt{14400+32\left(-448\right)}}{2\left(-8\right)}$$

Multiply 32 times -448.

$$x=\frac{-120±\sqrt{14400-14336}}{2\left(-8\right)}$$

Add 14400 to -14336.

$$x=\frac{-120±\sqrt{64}}{2\left(-8\right)}$$

Take the square root of 64.

$$x=\frac{-120±8}{2\left(-8\right)}$$

Multiply 2 times -8.

$$x=\frac{-120±8}{-16}$$

Now solve the equation x=\frac{-120±8}{-16} when ± is plus. Add -120 to 8.

$$x=-\frac{112}{-16}$$

Divide -112 by -16.

$$x=7$$

Now solve the equation x=\frac{-120±8}{-16} when ± is minus. Subtract 8 from -120.

$$x=-\frac{128}{-16}$$

Divide -128 by -16.

$$x=8$$

The equation is now solved.

$$x=7 x=8$$

Substitute 7 for x in the equation \sqrt{x+1}-\sqrt{9-x}=\sqrt{2x-12}.

$$\sqrt{7+1}-\sqrt{9-7}=\sqrt{2\times 7-12}$$

Simplify. The value x=7 satisfies the equation.

$$2^{\frac{1}{2}}=2^{\frac{1}{2}}$$

Substitute 8 for x in the equation \sqrt{x+1}-\sqrt{9-x}=\sqrt{2x-12}.

$$\sqrt{8+1}-\sqrt{9-8}=\sqrt{2\times 8-12}$$

Simplify. The value x=8 satisfies the equation.

$$2=2$$

List all solutions of \sqrt{x+1}-\sqrt{9-x}=\sqrt{2x-12}.

$$x=7 x=8$$