Solve sqrt{5-x}=x+1

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

Solve for x

$$x=1$$

Solution Steps

Square both sides of the equation.

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

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

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

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

$$5-x=x^{2}+2x+1$$

Subtract x^{2} from both sides.

$$5-x-x^{2}=2x+1$$

Subtract 2x from both sides.

$$5-x-x^{2}-2x=1$$

Combine -x and -2x to get -3x.

$$5-3x-x^{2}=1$$

Subtract 1 from both sides.

$$5-3x-x^{2}-1=0$$

Subtract 1 from 5 to get 4.

$$4-3x-x^{2}=0$$

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

$$-x^{2}-3x+4=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+4. To find a and b, set up a system to be solved.

$$a+b=-3 ab=-4=-4$$

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 -4.

$$1,-4 2,-2$$

Calculate the sum for each pair.

$$1-4=-3 2-2=0$$

The solution is the pair that gives sum -3.

$$a=1 b=-4$$

Rewrite -x^{2}-3x+4 as \left(-x^{2}+x\right)+\left(-4x+4\right).

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

Factor out x in the first and 4 in the second group.

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

Factor out common term -x+1 by using distributive property.

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

To find equation solutions, solve -x+1=0 and x+4=0.

$$x=1 x=-4$$

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

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

Simplify. The value x=1 satisfies the equation.

$$2=2$$

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

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

Simplify. The value x=-4 does not satisfy the equation because the left and the right hand side have opposite signs.

$$3=-3$$

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

$$x=1$$