Solve 3-sqrt{x-1}=sqrt{4x+5}

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

Solve for x

$$x=1$$

Solution Steps

Square both sides of the equation.

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

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

$$9-6\sqrt{x-1}+\left(\sqrt{x-1}\right)^{2}=\left(\sqrt{4x+5}\right)^{2}$$

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

$$9-6\sqrt{x-1}+x-1=\left(\sqrt{4x+5}\right)^{2}$$

Subtract 1 from 9 to get 8.

$$8-6\sqrt{x-1}+x=\left(\sqrt{4x+5}\right)^{2}$$

Calculate \sqrt{4x+5} to the power of 2 and get 4x+5.

$$8-6\sqrt{x-1}+x=4x+5$$

Subtract 8+x from both sides of the equation.

$$-6\sqrt{x-1}=4x+5-\left(8+x\right)$$

To find the opposite of 8+x, find the opposite of each term.

$$-6\sqrt{x-1}=4x+5-8-x$$

Subtract 8 from 5 to get -3.

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

Combine 4x and -x to get 3x.

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

Square both sides of the equation.

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

Expand \left(-6\sqrt{x-1}\right)^{2}.

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

Calculate -6 to the power of 2 and get 36.

$$36\left(\sqrt{x-1}\right)^{2}=\left(3x-3\right)^{2}$$

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

$$36\left(x-1\right)=\left(3x-3\right)^{2}$$

Use the distributive property to multiply 36 by x-1.

$$36x-36=\left(3x-3\right)^{2}$$

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

$$36x-36=9x^{2}-18x+9$$

Subtract 9x^{2} from both sides.

$$36x-36-9x^{2}=-18x+9$$

Add 18x to both sides.

$$36x-36-9x^{2}+18x=9$$

Combine 36x and 18x to get 54x.

$$54x-36-9x^{2}=9$$

Subtract 9 from both sides.

$$54x-36-9x^{2}-9=0$$

Subtract 9 from -36 to get -45.

$$54x-45-9x^{2}=0$$

Divide both sides by 9.

$$6x-5-x^{2}=0$$

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

$$-x^{2}+6x-5=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-5. To find a and b, set up a system to be solved.

$$a+b=6 ab=-\left(-5\right)=5$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=5 b=1$$

Rewrite -x^{2}+6x-5 as \left(-x^{2}+5x\right)+\left(x-5\right).

$$\left(-x^{2}+5x\right)+\left(x-5\right)$$

Factor out -x in -x^{2}+5x.

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

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

$$\left(x-5\right)\left(-x+1\right)$$

To find equation solutions, solve x-5=0 and -x+1=0.

$$x=5 x=1$$

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

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

Simplify. The value x=5 does not satisfy the equation.

$$1=5$$

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

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

Simplify. The value x=1 satisfies the equation.

$$3=3$$

Equation -\sqrt{x-1}+3=\sqrt{4x+5} has a unique solution.

$$x=1$$