Solve sqrt{2x-1}-sqrt{x-1}=sqrt{6-x}

$$\sqrt { 2 x - 1 } - \sqrt { x - 1 } = \sqrt { 6 - x }$$

Solve for x

$$x=5$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{2x-1}-\sqrt{x-1}\right)^{2}=\left(\sqrt{6-x}\right)^{2}$$

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

$$\left(\sqrt{2x-1}\right)^{2}-2\sqrt{2x-1}\sqrt{x-1}+\left(\sqrt{x-1}\right)^{2}=\left(\sqrt{6-x}\right)^{2}$$

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

$$2x-1-2\sqrt{2x-1}\sqrt{x-1}+\left(\sqrt{x-1}\right)^{2}=\left(\sqrt{6-x}\right)^{2}$$

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

$$2x-1-2\sqrt{2x-1}\sqrt{x-1}+x-1=\left(\sqrt{6-x}\right)^{2}$$

Combine 2x and x to get 3x.

$$3x-1-2\sqrt{2x-1}\sqrt{x-1}-1=\left(\sqrt{6-x}\right)^{2}$$

Subtract 1 from -1 to get -2.

$$3x-2-2\sqrt{2x-1}\sqrt{x-1}=\left(\sqrt{6-x}\right)^{2}$$

Calculate \sqrt{6-x} to the power of 2 and get 6-x.

$$3x-2-2\sqrt{2x-1}\sqrt{x-1}=6-x$$

Subtract 3x-2 from both sides of the equation.

$$-2\sqrt{2x-1}\sqrt{x-1}=6-x-\left(3x-2\right)$$

To find the opposite of 3x-2, find the opposite of each term.

$$-2\sqrt{2x-1}\sqrt{x-1}=6-x-3x+2$$

Combine -x and -3x to get -4x.

$$-2\sqrt{2x-1}\sqrt{x-1}=6-4x+2$$

Add 6 and 2 to get 8.

$$-2\sqrt{2x-1}\sqrt{x-1}=8-4x$$

Square both sides of the equation.

$$\left(-2\sqrt{2x-1}\sqrt{x-1}\right)^{2}=\left(8-4x\right)^{2}$$

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

$$\left(-2\right)^{2}\left(\sqrt{2x-1}\right)^{2}\left(\sqrt{x-1}\right)^{2}=\left(8-4x\right)^{2}$$

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

$$4\left(\sqrt{2x-1}\right)^{2}\left(\sqrt{x-1}\right)^{2}=\left(8-4x\right)^{2}$$

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

$$4\left(2x-1\right)\left(\sqrt{x-1}\right)^{2}=\left(8-4x\right)^{2}$$

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

$$4\left(2x-1\right)\left(x-1\right)=\left(8-4x\right)^{2}$$

Use the distributive property to multiply 4 by 2x-1.

$$\left(8x-4\right)\left(x-1\right)=\left(8-4x\right)^{2}$$

Apply the distributive property by multiplying each term of 8x-4 by each term of x-1.

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

Combine -8x and -4x to get -12x.

$$8x^{2}-12x+4=\left(8-4x\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(8-4x\right)^{2}.

$$8x^{2}-12x+4=64-64x+16x^{2}$$

Subtract 64 from both sides.

$$8x^{2}-12x+4-64=-64x+16x^{2}$$

Subtract 64 from 4 to get -60.

$$8x^{2}-12x-60=-64x+16x^{2}$$

Add 64x to both sides.

$$8x^{2}-12x-60+64x=16x^{2}$$

Combine -12x and 64x to get 52x.

$$8x^{2}+52x-60=16x^{2}$$

Subtract 16x^{2} from both sides.

$$8x^{2}+52x-60-16x^{2}=0$$

Combine 8x^{2} and -16x^{2} to get -8x^{2}.

$$-8x^{2}+52x-60=0$$

Divide both sides by 4.

$$-2x^{2}+13x-15=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -2x^{2}+ax+bx-15. To find a and b, set up a system to be solved.

$$a+b=13 ab=-2\left(-15\right)=30$$

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 30.

$$1,30 2,15 3,10 5,6$$

Calculate the sum for each pair.

$$1+30=31 2+15=17 3+10=13 5+6=11$$

The solution is the pair that gives sum 13.

$$a=10 b=3$$

Rewrite -2x^{2}+13x-15 as \left(-2x^{2}+10x\right)+\left(3x-15\right).

$$\left(-2x^{2}+10x\right)+\left(3x-15\right)$$

Factor out 2x in the first and -3 in the second group.

$$2x\left(-x+5\right)-3\left(-x+5\right)$$

Factor out common term -x+5 by using distributive property.

$$\left(-x+5\right)\left(2x-3\right)$$

To find equation solutions, solve -x+5=0 and 2x-3=0.

$$x=5 x=\frac{3}{2}$$

Substitute 5 for x in the equation \sqrt{2x-1}-\sqrt{x-1}=\sqrt{6-x}.

$$\sqrt{2\times 5-1}-\sqrt{5-1}=\sqrt{6-5}$$

Simplify. The value x=5 satisfies the equation.

$$1=1$$

Substitute \frac{3}{2} for x in the equation \sqrt{2x-1}-\sqrt{x-1}=\sqrt{6-x}.

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

Simplify. The value x=\frac{3}{2} does not satisfy the equation.

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

Substitute 5 for x in the equation \sqrt{2x-1}-\sqrt{x-1}=\sqrt{6-x}.

$$\sqrt{2\times 5-1}-\sqrt{5-1}=\sqrt{6-5}$$

Simplify. The value x=5 satisfies the equation.

$$1=1$$

Equation \sqrt{2x-1}-\sqrt{x-1}=\sqrt{6-x} has a unique solution.

$$x=5$$