Solve 4sqrt{x}-2=sqrt{5-x}

$$4 \sqrt { x } - 2 = \sqrt { 5 - x }$$

Solve for x

$$x=1$$

Solution Steps

Square both sides of the equation.

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

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

$$16\left(\sqrt{x}\right)^{2}-16\sqrt{x}+4=\left(\sqrt{5-x}\right)^{2}$$

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

$$16x-16\sqrt{x}+4=\left(\sqrt{5-x}\right)^{2}$$

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

$$16x-16\sqrt{x}+4=5-x$$

Subtract 16x+4 from both sides of the equation.

$$-16\sqrt{x}=5-x-\left(16x+4\right)$$

To find the opposite of 16x+4, find the opposite of each term.

$$-16\sqrt{x}=5-x-16x-4$$

Combine -x and -16x to get -17x.

$$-16\sqrt{x}=5-17x-4$$

Subtract 4 from 5 to get 1.

$$-16\sqrt{x}=1-17x$$

Square both sides of the equation.

$$\left(-16\sqrt{x}\right)^{2}=\left(1-17x\right)^{2}$$

Expand \left(-16\sqrt{x}\right)^{2}.

$$\left(-16\right)^{2}\left(\sqrt{x}\right)^{2}=\left(1-17x\right)^{2}$$

Calculate -16 to the power of 2 and get 256.

$$256\left(\sqrt{x}\right)^{2}=\left(1-17x\right)^{2}$$

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

$$256x=\left(1-17x\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(1-17x\right)^{2}.

$$256x=1-34x+289x^{2}$$

Subtract 1 from both sides.

$$256x-1=-34x+289x^{2}$$

Add 34x to both sides.

$$256x-1+34x=289x^{2}$$

Combine 256x and 34x to get 290x.

$$290x-1=289x^{2}$$

Subtract 289x^{2} from both sides.

$$290x-1-289x^{2}=0$$

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

$$-289x^{2}+290x-1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -289x^{2}+ax+bx-1. To find a and b, set up a system to be solved.

$$a+b=290 ab=-289\left(-1\right)=289$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 289.

$$1,289 17,17$$

Calculate the sum for each pair.

$$1+289=290 17+17=34$$

The solution is the pair that gives sum 290.

$$a=289 b=1$$

Rewrite -289x^{2}+290x-1 as \left(-289x^{2}+289x\right)+\left(x-1\right).

$$\left(-289x^{2}+289x\right)+\left(x-1\right)$$

Factor out 289x in the first and -1 in the second group.

$$289x\left(-x+1\right)-\left(-x+1\right)$$

Factor out common term -x+1 by using distributive property.

$$\left(-x+1\right)\left(289x-1\right)$$

To find equation solutions, solve -x+1=0 and 289x-1=0.

$$x=1 x=\frac{1}{289}$$

Substitute 1 for x in the equation 4\sqrt{x}-2=\sqrt{5-x}.

$$4\sqrt{1}-2=\sqrt{5-1}$$

Simplify. The value x=1 satisfies the equation.

$$2=2$$

Substitute \frac{1}{289} for x in the equation 4\sqrt{x}-2=\sqrt{5-x}.

$$4\sqrt{\frac{1}{289}}-2=\sqrt{5-\frac{1}{289}}$$

Simplify. The value x=\frac{1}{289} does not satisfy the equation because the left and the right hand side have opposite signs.

$$-\frac{30}{17}=\frac{38}{17}$$

Substitute 1 for x in the equation 4\sqrt{x}-2=\sqrt{5-x}.

$$4\sqrt{1}-2=\sqrt{5-1}$$

Simplify. The value x=1 satisfies the equation.

$$2=2$$

Equation 4\sqrt{x}-2=\sqrt{5-x} has a unique solution.

$$x=1$$