Solve sqrt{3x-4}-sqrt{x-4}=4

$$\sqrt { 3 x - 4 } - \sqrt { x - 4 } = 4$$

Solve for x

$$x=8\sqrt{2}+16\approx 27.313708499$$

Solution Steps

Subtract -\sqrt{x-4} from both sides of the equation.

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

Square both sides of the equation.

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

Calculate \sqrt{3x-4} to the power of 2 and get 3x-4.

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

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

$$3x-4=16+8\sqrt{x-4}+\left(\sqrt{x-4}\right)^{2}$$

Calculate \sqrt{x-4} to the power of 2 and get x-4.

$$3x-4=16+8\sqrt{x-4}+x-4$$

Subtract 4 from 16 to get 12.

$$3x-4=12+8\sqrt{x-4}+x$$

Subtract 12+x from both sides of the equation.

$$3x-4-\left(12+x\right)=8\sqrt{x-4}$$

To find the opposite of 12+x, find the opposite of each term.

$$3x-4-12-x=8\sqrt{x-4}$$

Subtract 12 from -4 to get -16.

$$3x-16-x=8\sqrt{x-4}$$

Combine 3x and -x to get 2x.

$$2x-16=8\sqrt{x-4}$$

Square both sides of the equation.

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-16\right)^{2}.

$$4x^{2}-64x+256=\left(8\sqrt{x-4}\right)^{2}$$

Expand \left(8\sqrt{x-4}\right)^{2}.

$$4x^{2}-64x+256=8^{2}\left(\sqrt{x-4}\right)^{2}$$

Calculate 8 to the power of 2 and get 64.

$$4x^{2}-64x+256=64\left(\sqrt{x-4}\right)^{2}$$

Calculate \sqrt{x-4} to the power of 2 and get x-4.

$$4x^{2}-64x+256=64\left(x-4\right)$$

Use the distributive property to multiply 64 by x-4.

$$4x^{2}-64x+256=64x-256$$

Subtract 64x from both sides.

$$4x^{2}-64x+256-64x=-256$$

Combine -64x and -64x to get -128x.

$$4x^{2}-128x+256=-256$$

Add 256 to both sides.

$$4x^{2}-128x+256+256=0$$

Add 256 and 256 to get 512.

$$4x^{2}-128x+512=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -128 for b, and 512 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-128\right)±\sqrt{\left(-128\right)^{2}-4\times 4\times 512}}{2\times 4}$$

Square -128.

$$x=\frac{-\left(-128\right)±\sqrt{16384-4\times 4\times 512}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-128\right)±\sqrt{16384-16\times 512}}{2\times 4}$$

Multiply -16 times 512.

$$x=\frac{-\left(-128\right)±\sqrt{16384-8192}}{2\times 4}$$

Add 16384 to -8192.

$$x=\frac{-\left(-128\right)±\sqrt{8192}}{2\times 4}$$

Take the square root of 8192.

$$x=\frac{-\left(-128\right)±64\sqrt{2}}{2\times 4}$$

The opposite of -128 is 128.

$$x=\frac{128±64\sqrt{2}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{128±64\sqrt{2}}{8}$$

Now solve the equation x=\frac{128±64\sqrt{2}}{8} when ± is plus. Add 128 to 64\sqrt{2}.

$$x=\frac{64\sqrt{2}+128}{8}$$

Divide 128+64\sqrt{2} by 8.

$$x=8\sqrt{2}+16$$

Now solve the equation x=\frac{128±64\sqrt{2}}{8} when ± is minus. Subtract 64\sqrt{2} from 128.

$$x=\frac{128-64\sqrt{2}}{8}$$

Divide 128-64\sqrt{2} by 8.

$$x=16-8\sqrt{2}$$

The equation is now solved.

$$x=8\sqrt{2}+16 x=16-8\sqrt{2}$$

Substitute 8\sqrt{2}+16 for x in the equation \sqrt{3x-4}-\sqrt{x-4}=4.

$$\sqrt{3\left(8\sqrt{2}+16\right)-4}-\sqrt{8\sqrt{2}+16-4}=4$$

Simplify. The value x=8\sqrt{2}+16 satisfies the equation.

$$4=4$$

Substitute 16-8\sqrt{2} for x in the equation \sqrt{3x-4}-\sqrt{x-4}=4.

$$\sqrt{3\left(16-8\sqrt{2}\right)-4}-\sqrt{16-8\sqrt{2}-4}=4$$

Simplify. The value x=16-8\sqrt{2} does not satisfy the equation.

$$8-4\times 2^{\frac{1}{2}}=4$$

Substitute 8\sqrt{2}+16 for x in the equation \sqrt{3x-4}-\sqrt{x-4}=4.

$$\sqrt{3\left(8\sqrt{2}+16\right)-4}-\sqrt{8\sqrt{2}+16-4}=4$$

Simplify. The value x=8\sqrt{2}+16 satisfies the equation.

$$4=4$$

Equation \sqrt{3x-4}=\sqrt{x-4}+4 has a unique solution.

$$x=8\sqrt{2}+16$$