Solve sqrt{x+4}+sqrt{2x-1}=3*sqrt{x-1}

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

Solve for x

$$x=5$$

Solution Steps

Square both sides of the equation.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{x+4}+\sqrt{2x-1}\right)^{2}.

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

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

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

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

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

Combine x and 2x to get 3x.

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

Subtract 1 from 4 to get 3.

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

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

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

Calculate 3 to the power of 2 and get 9.

$$3x+3+2\sqrt{x+4}\sqrt{2x-1}=9\left(\sqrt{x-1}\right)^{2}$$

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

$$3x+3+2\sqrt{x+4}\sqrt{2x-1}=9\left(x-1\right)$$

Use the distributive property to multiply 9 by x-1.

$$3x+3+2\sqrt{x+4}\sqrt{2x-1}=9x-9$$

Subtract 3x+3 from both sides of the equation.

$$2\sqrt{x+4}\sqrt{2x-1}=9x-9-\left(3x+3\right)$$

To find the opposite of 3x+3, find the opposite of each term.

$$2\sqrt{x+4}\sqrt{2x-1}=9x-9-3x-3$$

Combine 9x and -3x to get 6x.

$$2\sqrt{x+4}\sqrt{2x-1}=6x-9-3$$

Subtract 3 from -9 to get -12.

$$2\sqrt{x+4}\sqrt{2x-1}=6x-12$$

Square both sides of the equation.

$$\left(2\sqrt{x+4}\sqrt{2x-1}\right)^{2}=\left(6x-12\right)^{2}$$

Expand \left(2\sqrt{x+4}\sqrt{2x-1}\right)^{2}.

$$2^{2}\left(\sqrt{x+4}\right)^{2}\left(\sqrt{2x-1}\right)^{2}=\left(6x-12\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$4\left(\sqrt{x+4}\right)^{2}\left(\sqrt{2x-1}\right)^{2}=\left(6x-12\right)^{2}$$

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

$$4\left(x+4\right)\left(\sqrt{2x-1}\right)^{2}=\left(6x-12\right)^{2}$$

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

$$4\left(x+4\right)\left(2x-1\right)=\left(6x-12\right)^{2}$$

Use the distributive property to multiply 4 by x+4.

$$\left(4x+16\right)\left(2x-1\right)=\left(6x-12\right)^{2}$$

Apply the distributive property by multiplying each term of 4x+16 by each term of 2x-1.

$$8x^{2}-4x+32x-16=\left(6x-12\right)^{2}$$

Combine -4x and 32x to get 28x.

$$8x^{2}+28x-16=\left(6x-12\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(6x-12\right)^{2}.

$$8x^{2}+28x-16=36x^{2}-144x+144$$

Subtract 36x^{2} from both sides.

$$8x^{2}+28x-16-36x^{2}=-144x+144$$

Combine 8x^{2} and -36x^{2} to get -28x^{2}.

$$-28x^{2}+28x-16=-144x+144$$

Add 144x to both sides.

$$-28x^{2}+28x-16+144x=144$$

Combine 28x and 144x to get 172x.

$$-28x^{2}+172x-16=144$$

Subtract 144 from both sides.

$$-28x^{2}+172x-16-144=0$$

Subtract 144 from -16 to get -160.

$$-28x^{2}+172x-160=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -28 for a, 172 for b, and -160 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-172±\sqrt{172^{2}-4\left(-28\right)\left(-160\right)}}{2\left(-28\right)}$$

Square 172.

$$x=\frac{-172±\sqrt{29584-4\left(-28\right)\left(-160\right)}}{2\left(-28\right)}$$

Multiply -4 times -28.

$$x=\frac{-172±\sqrt{29584+112\left(-160\right)}}{2\left(-28\right)}$$

Multiply 112 times -160.

$$x=\frac{-172±\sqrt{29584-17920}}{2\left(-28\right)}$$

Add 29584 to -17920.

$$x=\frac{-172±\sqrt{11664}}{2\left(-28\right)}$$

Take the square root of 11664.

$$x=\frac{-172±108}{2\left(-28\right)}$$

Multiply 2 times -28.

$$x=\frac{-172±108}{-56}$$

Now solve the equation x=\frac{-172±108}{-56} when ± is plus. Add -172 to 108.

$$x=-\frac{64}{-56}$$

Reduce the fraction \frac{-64}{-56} to lowest terms by extracting and canceling out 8.

$$x=\frac{8}{7}$$

Now solve the equation x=\frac{-172±108}{-56} when ± is minus. Subtract 108 from -172.

$$x=-\frac{280}{-56}$$

Divide -280 by -56.

$$x=5$$

The equation is now solved.

$$x=\frac{8}{7} x=5$$

Substitute \frac{8}{7} for x in the equation \sqrt{x+4}+\sqrt{2x-1}=3\sqrt{x-1}.

$$\sqrt{\frac{8}{7}+4}+\sqrt{2\times \frac{8}{7}-1}=3\sqrt{\frac{8}{7}-1}$$

Simplify. The value x=\frac{8}{7} does not satisfy the equation.

$$\frac{9}{7}\times 7^{\frac{1}{2}}=\frac{3}{7}\times 7^{\frac{1}{2}}$$

Substitute 5 for x in the equation \sqrt{x+4}+\sqrt{2x-1}=3\sqrt{x-1}.

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

Simplify. The value x=5 satisfies the equation.

$$6=6$$

Equation \sqrt{x+4}+\sqrt{2x-1}=3\sqrt{x-1} has a unique solution.

$$x=5$$