Solve sqrt{5x-1}-sqrt{3-x}=sqrt{2x}

$$\sqrt { 5 x - 1 } - \sqrt { 3 - x } = \sqrt { 2 x }$$

Solve for x

$$x=2$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{5x-1}-\sqrt{3-x}\right)^{2}=\left(\sqrt{2x}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{5x-1}-\sqrt{3-x}\right)^{2}.

$$\left(\sqrt{5x-1}\right)^{2}-2\sqrt{5x-1}\sqrt{3-x}+\left(\sqrt{3-x}\right)^{2}=\left(\sqrt{2x}\right)^{2}$$

Calculate \sqrt{5x-1} to the power of 2 and get 5x-1.

$$5x-1-2\sqrt{5x-1}\sqrt{3-x}+\left(\sqrt{3-x}\right)^{2}=\left(\sqrt{2x}\right)^{2}$$

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

$$5x-1-2\sqrt{5x-1}\sqrt{3-x}+3-x=\left(\sqrt{2x}\right)^{2}$$

Add -1 and 3 to get 2.

$$5x+2-2\sqrt{5x-1}\sqrt{3-x}-x=\left(\sqrt{2x}\right)^{2}$$

Combine 5x and -x to get 4x.

$$4x+2-2\sqrt{5x-1}\sqrt{3-x}=\left(\sqrt{2x}\right)^{2}$$

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

$$4x+2-2\sqrt{5x-1}\sqrt{3-x}=2x$$

Subtract 4x+2 from both sides of the equation.

$$-2\sqrt{5x-1}\sqrt{3-x}=2x-\left(4x+2\right)$$

To find the opposite of 4x+2, find the opposite of each term.

$$-2\sqrt{5x-1}\sqrt{3-x}=2x-4x-2$$

Combine 2x and -4x to get -2x.

$$-2\sqrt{5x-1}\sqrt{3-x}=-2x-2$$

Square both sides of the equation.

$$\left(-2\sqrt{5x-1}\sqrt{3-x}\right)^{2}=\left(-2x-2\right)^{2}$$

Expand \left(-2\sqrt{5x-1}\sqrt{3-x}\right)^{2}.

$$\left(-2\right)^{2}\left(\sqrt{5x-1}\right)^{2}\left(\sqrt{3-x}\right)^{2}=\left(-2x-2\right)^{2}$$

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

$$4\left(\sqrt{5x-1}\right)^{2}\left(\sqrt{3-x}\right)^{2}=\left(-2x-2\right)^{2}$$

Calculate \sqrt{5x-1} to the power of 2 and get 5x-1.

$$4\left(5x-1\right)\left(\sqrt{3-x}\right)^{2}=\left(-2x-2\right)^{2}$$

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

$$4\left(5x-1\right)\left(3-x\right)=\left(-2x-2\right)^{2}$$

Use the distributive property to multiply 4 by 5x-1.

$$\left(20x-4\right)\left(3-x\right)=\left(-2x-2\right)^{2}$$

Apply the distributive property by multiplying each term of 20x-4 by each term of 3-x.

$$60x-20x^{2}-12+4x=\left(-2x-2\right)^{2}$$

Combine 60x and 4x to get 64x.

$$64x-20x^{2}-12=\left(-2x-2\right)^{2}$$

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

$$64x-20x^{2}-12=4x^{2}+8x+4$$

Subtract 4x^{2} from both sides.

$$64x-20x^{2}-12-4x^{2}=8x+4$$

Combine -20x^{2} and -4x^{2} to get -24x^{2}.

$$64x-24x^{2}-12=8x+4$$

Subtract 8x from both sides.

$$64x-24x^{2}-12-8x=4$$

Combine 64x and -8x to get 56x.

$$56x-24x^{2}-12=4$$

Subtract 4 from both sides.

$$56x-24x^{2}-12-4=0$$

Subtract 4 from -12 to get -16.

$$56x-24x^{2}-16=0$$

Divide both sides by 8.

$$7x-3x^{2}-2=0$$

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

$$-3x^{2}+7x-2=0$$

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

$$a+b=7 ab=-3\left(-2\right)=6$$

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 6.

$$1,6 2,3$$

Calculate the sum for each pair.

$$1+6=7 2+3=5$$

The solution is the pair that gives sum 7.

$$a=6 b=1$$

Rewrite -3x^{2}+7x-2 as \left(-3x^{2}+6x\right)+\left(x-2\right).

$$\left(-3x^{2}+6x\right)+\left(x-2\right)$$

Factor out 3x in the first and -1 in the second group.

$$3x\left(-x+2\right)-\left(-x+2\right)$$

Factor out common term -x+2 by using distributive property.

$$\left(-x+2\right)\left(3x-1\right)$$

To find equation solutions, solve -x+2=0 and 3x-1=0.

$$x=2 x=\frac{1}{3}$$

Substitute 2 for x in the equation \sqrt{5x-1}-\sqrt{3-x}=\sqrt{2x}.

$$\sqrt{5\times 2-1}-\sqrt{3-2}=\sqrt{2\times 2}$$

Simplify. The value x=2 satisfies the equation.

$$2=2$$

Substitute \frac{1}{3} for x in the equation \sqrt{5x-1}-\sqrt{3-x}=\sqrt{2x}.

$$\sqrt{5\times \frac{1}{3}-1}-\sqrt{3-\frac{1}{3}}=\sqrt{2\times \frac{1}{3}}$$

Simplify. The value x=\frac{1}{3} does not satisfy the equation because the left and the right hand side have opposite signs.

$$-\frac{1}{3}\times 6^{\frac{1}{2}}=\frac{1}{3}\times 6^{\frac{1}{2}}$$

Substitute 2 for x in the equation \sqrt{5x-1}-\sqrt{3-x}=\sqrt{2x}.

$$\sqrt{5\times 2-1}-\sqrt{3-2}=\sqrt{2\times 2}$$

Simplify. The value x=2 satisfies the equation.

$$2=2$$

Equation \sqrt{5x-1}-\sqrt{3-x}=\sqrt{2x} has a unique solution.

$$x=2$$