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

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

Solve for x

$$x=5$$

Solution Steps

Subtract \sqrt{x-1} from both sides of the equation.

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

Square both sides of the equation.

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

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

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5-\sqrt{x-1}\right)^{2}.

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

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

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

Subtract 1 from 25 to get 24.

$$x+4=24-10\sqrt{x-1}+x$$

Add 10\sqrt{x-1} to both sides.

$$x+4+10\sqrt{x-1}=24+x$$

Subtract x from both sides.

$$x+4+10\sqrt{x-1}-x=24$$

Combine x and -x to get 0.

$$4+10\sqrt{x-1}=24$$

Subtract 4 from both sides.

$$10\sqrt{x-1}=24-4$$

Subtract 4 from 24 to get 20.

$$10\sqrt{x-1}=20$$

Divide both sides by 10.

$$\sqrt{x-1}=\frac{20}{10}$$

Divide 20 by 10 to get 2.

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

Square both sides of the equation.

$$x-1=4$$

Add 1 to both sides of the equation.

$$x-1-\left(-1\right)=4-\left(-1\right)$$

Subtracting -1 from itself leaves 0.

$$x=4-\left(-1\right)$$

Subtract -1 from 4.

$$x=5$$

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

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

Simplify. The value x=5 satisfies the equation.

$$5=5$$

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

$$x=5$$