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

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

Solve for x

$$x=5$$

Solution Steps

Square both sides of the equation.

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

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

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

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

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

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

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

Combine x and x to get 2x.

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

Subtract 4 from -1 to get -5.

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

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

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

Subtract 2x-5 from both sides of the equation.

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

To find the opposite of 2x-5, find the opposite of each term.

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

Combine x and -2x to get -x.

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

Add 4 and 5 to get 9.

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

Square both sides of the equation.

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

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

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

Calculate 2 to the power of 2 and get 4.

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

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

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

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

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

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

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

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

$$4x^{2}-16x-4x+16=\left(-x+9\right)^{2}$$

Combine -16x and -4x to get -20x.

$$4x^{2}-20x+16=\left(-x+9\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(-x+9\right)^{2}.

$$4x^{2}-20x+16=x^{2}-18x+81$$

Subtract x^{2} from both sides.

$$4x^{2}-20x+16-x^{2}=-18x+81$$

Combine 4x^{2} and -x^{2} to get 3x^{2}.

$$3x^{2}-20x+16=-18x+81$$

Add 18x to both sides.

$$3x^{2}-20x+16+18x=81$$

Combine -20x and 18x to get -2x.

$$3x^{2}-2x+16=81$$

Subtract 81 from both sides.

$$3x^{2}-2x+16-81=0$$

Subtract 81 from 16 to get -65.

$$3x^{2}-2x-65=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx-65. To find a and b, set up a system to be solved.

$$a+b=-2 ab=3\left(-65\right)=-195$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -195.

$$1,-195 3,-65 5,-39 13,-15$$

Calculate the sum for each pair.

$$1-195=-194 3-65=-62 5-39=-34 13-15=-2$$

The solution is the pair that gives sum -2.

$$a=-15 b=13$$

Rewrite 3x^{2}-2x-65 as \left(3x^{2}-15x\right)+\left(13x-65\right).

$$\left(3x^{2}-15x\right)+\left(13x-65\right)$$

Factor out 3x in the first and 13 in the second group.

$$3x\left(x-5\right)+13\left(x-5\right)$$

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

$$\left(x-5\right)\left(3x+13\right)$$

To find equation solutions, solve x-5=0 and 3x+13=0.

$$x=5 x=-\frac{13}{3}$$

Substitute -\frac{13}{3} for x in the equation \sqrt{x-1}+\sqrt{x-4}=\sqrt{x+4}. The expression \sqrt{-\frac{13}{3}-1} is undefined because the radicand cannot be negative.

$$\sqrt{-\frac{13}{3}-1}+\sqrt{-\frac{13}{3}-4}=\sqrt{-\frac{13}{3}+4}$$

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

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

Simplify. The value x=5 satisfies the equation.

$$3=3$$

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

$$x=5$$