Solve sqrt{41-x}=x+1

$$\sqrt { 41 - x } = x + 1$$

Solve for x

$$x=5$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{41-x}\right)^{2}=\left(x+1\right)^{2}$$

Calculate \sqrt{41-x} to the power of 2 and get 41-x.

$$41-x=\left(x+1\right)^{2}$$

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

$$41-x=x^{2}+2x+1$$

Subtract x^{2} from both sides.

$$41-x-x^{2}=2x+1$$

Subtract 2x from both sides.

$$41-x-x^{2}-2x=1$$

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

$$41-3x-x^{2}=1$$

Subtract 1 from both sides.

$$41-3x-x^{2}-1=0$$

Subtract 1 from 41 to get 40.

$$40-3x-x^{2}=0$$

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

$$-x^{2}-3x+40=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+40. To find a and b, set up a system to be solved.

$$a+b=-3 ab=-40=-40$$

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 -40.

$$1,-40 2,-20 4,-10 5,-8$$

Calculate the sum for each pair.

$$1-40=-39 2-20=-18 4-10=-6 5-8=-3$$

The solution is the pair that gives sum -3.

$$a=5 b=-8$$

Rewrite -x^{2}-3x+40 as \left(-x^{2}+5x\right)+\left(-8x+40\right).

$$\left(-x^{2}+5x\right)+\left(-8x+40\right)$$

Factor out x in the first and 8 in the second group.

$$x\left(-x+5\right)+8\left(-x+5\right)$$

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

$$\left(-x+5\right)\left(x+8\right)$$

To find equation solutions, solve -x+5=0 and x+8=0.

$$x=5 x=-8$$

Substitute 5 for x in the equation \sqrt{41-x}=x+1.

$$\sqrt{41-5}=5+1$$

Simplify. The value x=5 satisfies the equation.

$$6=6$$

Substitute -8 for x in the equation \sqrt{41-x}=x+1.

$$\sqrt{41-\left(-8\right)}=-8+1$$

Simplify. The value x=-8 does not satisfy the equation because the left and the right hand side have opposite signs.

$$7=-7$$

Equation \sqrt{41-x}=x+1 has a unique solution.

$$x=5$$