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

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

Solve for x

$$x=5$$

Solution Steps

Square both sides of the equation.

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

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

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5-\sqrt{x+4}\right)^{2}.

$$x-1=25-10\sqrt{x+4}+\left(\sqrt{x+4}\right)^{2}$$

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

$$x-1=25-10\sqrt{x+4}+x+4$$

Add 25 and 4 to get 29.

$$x-1=29-10\sqrt{x+4}+x$$

Add 10\sqrt{x+4} to both sides.

$$x-1+10\sqrt{x+4}=29+x$$

Subtract x from both sides.

$$x-1+10\sqrt{x+4}-x=29$$

Combine x and -x to get 0.

$$-1+10\sqrt{x+4}=29$$

Add 1 to both sides.

$$10\sqrt{x+4}=29+1$$

Add 29 and 1 to get 30.

$$10\sqrt{x+4}=30$$

Divide both sides by 10.

$$\sqrt{x+4}=\frac{30}{10}$$

Divide 30 by 10 to get 3.

$$\sqrt{x+4}=3$$

Square both sides of the equation.

$$x+4=9$$

Subtract 4 from both sides of the equation.

$$x+4-4=9-4$$

Subtracting 4 from itself leaves 0.

$$x=9-4$$

Subtract 4 from 9.

$$x=5$$

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

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

Simplify. The value x=5 satisfies the equation.

$$2=2$$

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

$$x=5$$