Solve sqrt{3x+4}-sqrt{x-4}=2sqrt{x}

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

Solve for x

$$x=4$$

Solution Steps

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

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

Square both sides of the equation.

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

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

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

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

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

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

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

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

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

Combine 4x and x to get 5x.

$$3x+4=5x+4\sqrt{x}\sqrt{x-4}-4$$

Subtract 5x-4 from both sides of the equation.

$$3x+4-\left(5x-4\right)=4\sqrt{x}\sqrt{x-4}$$

To find the opposite of 5x-4, find the opposite of each term.

$$3x+4-5x+4=4\sqrt{x}\sqrt{x-4}$$

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

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

Add 4 and 4 to get 8.

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

Square both sides of the equation.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-2x+8\right)^{2}.

$$4x^{2}-32x+64=\left(4\sqrt{x}\sqrt{x-4}\right)^{2}$$

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

$$4x^{2}-32x+64=4^{2}\left(\sqrt{x}\right)^{2}\left(\sqrt{x-4}\right)^{2}$$

Calculate 4 to the power of 2 and get 16.

$$4x^{2}-32x+64=16\left(\sqrt{x}\right)^{2}\left(\sqrt{x-4}\right)^{2}$$

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

$$4x^{2}-32x+64=16x\left(\sqrt{x-4}\right)^{2}$$

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

$$4x^{2}-32x+64=16x\left(x-4\right)$$

Use the distributive property to multiply 16x by x-4.

$$4x^{2}-32x+64=16x^{2}-64x$$

Subtract 16x^{2} from both sides.

$$4x^{2}-32x+64-16x^{2}=-64x$$

Combine 4x^{2} and -16x^{2} to get -12x^{2}.

$$-12x^{2}-32x+64=-64x$$

Add 64x to both sides.

$$-12x^{2}-32x+64+64x=0$$

Combine -32x and 64x to get 32x.

$$-12x^{2}+32x+64=0$$

Divide both sides by 4.

$$-3x^{2}+8x+16=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -3x^{2}+ax+bx+16. To find a and b, set up a system to be solved.

$$a+b=8 ab=-3\times 16=-48$$

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 -48.

$$-1,48 -2,24 -3,16 -4,12 -6,8$$

Calculate the sum for each pair.

$$-1+48=47 -2+24=22 -3+16=13 -4+12=8 -6+8=2$$

The solution is the pair that gives sum 8.

$$a=12 b=-4$$

Rewrite -3x^{2}+8x+16 as \left(-3x^{2}+12x\right)+\left(-4x+16\right).

$$\left(-3x^{2}+12x\right)+\left(-4x+16\right)$$

Factor out 3x in the first and 4 in the second group.

$$3x\left(-x+4\right)+4\left(-x+4\right)$$

Factor out common term -x+4 by using distributive property.

$$\left(-x+4\right)\left(3x+4\right)$$

To find equation solutions, solve -x+4=0 and 3x+4=0.

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

Substitute -\frac{4}{3} for x in the equation \sqrt{3x+4}-\sqrt{x-4}=2\sqrt{x}. The expression \sqrt{-\frac{4}{3}-4} is undefined because the radicand cannot be negative.

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

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

$$\sqrt{3\times 4+4}-\sqrt{4-4}=2\sqrt{4}$$

Simplify. The value x=4 satisfies the equation.

$$4=4$$

Equation \sqrt{3x+4}=\sqrt{x-4}+2\sqrt{x} has a unique solution.

$$x=4$$