Solve x+2sqrt{x-5}=6

$$x + 2 \sqrt { x - 5 } = 6$$

Solve for x

$$x=8-2\sqrt{2}\approx 5.171572875$$

Solution Steps

Subtract x from both sides of the equation.

$$2\sqrt{x-5}=6-x$$

Square both sides of the equation.

$$\left(2\sqrt{x-5}\right)^{2}=\left(6-x\right)^{2}$$

Expand \left(2\sqrt{x-5}\right)^{2}.

$$2^{2}\left(\sqrt{x-5}\right)^{2}=\left(6-x\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$4\left(\sqrt{x-5}\right)^{2}=\left(6-x\right)^{2}$$

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

$$4\left(x-5\right)=\left(6-x\right)^{2}$$

Use the distributive property to multiply 4 by x-5.

$$4x-20=\left(6-x\right)^{2}$$

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

$$4x-20=36-12x+x^{2}$$

Subtract 36 from both sides.

$$4x-20-36=-12x+x^{2}$$

Subtract 36 from -20 to get -56.

$$4x-56=-12x+x^{2}$$

Add 12x to both sides.

$$4x-56+12x=x^{2}$$

Combine 4x and 12x to get 16x.

$$16x-56=x^{2}$$

Subtract x^{2} from both sides.

$$16x-56-x^{2}=0$$

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.

$$-x^{2}+16x-56=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 16 for b, and -56 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-16±\sqrt{16^{2}-4\left(-1\right)\left(-56\right)}}{2\left(-1\right)}$$

Square 16.

$$x=\frac{-16±\sqrt{256-4\left(-1\right)\left(-56\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-16±\sqrt{256+4\left(-56\right)}}{2\left(-1\right)}$$

Multiply 4 times -56.

$$x=\frac{-16±\sqrt{256-224}}{2\left(-1\right)}$$

Add 256 to -224.

$$x=\frac{-16±\sqrt{32}}{2\left(-1\right)}$$

Take the square root of 32.

$$x=\frac{-16±4\sqrt{2}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-16±4\sqrt{2}}{-2}$$

Now solve the equation x=\frac{-16±4\sqrt{2}}{-2} when ± is plus. Add -16 to 4\sqrt{2}.

$$x=\frac{4\sqrt{2}-16}{-2}$$

Divide -16+4\sqrt{2} by -2.

$$x=8-2\sqrt{2}$$

Now solve the equation x=\frac{-16±4\sqrt{2}}{-2} when ± is minus. Subtract 4\sqrt{2} from -16.

$$x=\frac{-4\sqrt{2}-16}{-2}$$

Divide -16-4\sqrt{2} by -2.

$$x=2\sqrt{2}+8$$

The equation is now solved.

$$x=8-2\sqrt{2} x=2\sqrt{2}+8$$

Substitute 8-2\sqrt{2} for x in the equation x+2\sqrt{x-5}=6.

$$8-2\sqrt{2}+2\sqrt{8-2\sqrt{2}-5}=6$$

Simplify. The value x=8-2\sqrt{2} satisfies the equation.

$$6=6$$

Substitute 2\sqrt{2}+8 for x in the equation x+2\sqrt{x-5}=6.

$$2\sqrt{2}+8+2\sqrt{2\sqrt{2}+8-5}=6$$

Simplify. The value x=2\sqrt{2}+8 does not satisfy the equation.

$$4\times 2^{\frac{1}{2}}+10=6$$

Equation 2\sqrt{x-5}=6-x has a unique solution.

$$x=8-2\sqrt{2}$$