Solve sqrt{31-x}=x-1

$$\sqrt { 31 - x } = x - 1$$

Solve for x

$$x=6$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{31-x}\right)^{2}=\left(x-1\right)^{2}$$

Calculate \sqrt{31-x} to the power of 2 and get 31-x.

$$31-x=\left(x-1\right)^{2}$$

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

$$31-x=x^{2}-2x+1$$

Subtract x^{2} from both sides.

$$31-x-x^{2}=-2x+1$$

Add 2x to both sides.

$$31-x-x^{2}+2x=1$$

Combine -x and 2x to get x.

$$31+x-x^{2}=1$$

Subtract 1 from both sides.

$$31+x-x^{2}-1=0$$

Subtract 1 from 31 to get 30.

$$30+x-x^{2}=0$$

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

$$-x^{2}+x+30=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+30. To find a and b, set up a system to be solved.

$$a+b=1 ab=-30=-30$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -30.

$$-1,30 -2,15 -3,10 -5,6$$

Calculate the sum for each pair.

$$-1+30=29 -2+15=13 -3+10=7 -5+6=1$$

The solution is the pair that gives sum 1.

$$a=6 b=-5$$

Rewrite -x^{2}+x+30 as \left(-x^{2}+6x\right)+\left(-5x+30\right).

$$\left(-x^{2}+6x\right)+\left(-5x+30\right)$$

Factor out -x in the first and -5 in the second group.

$$-x\left(x-6\right)-5\left(x-6\right)$$

Factor out common term x-6 by using distributive property.

$$\left(x-6\right)\left(-x-5\right)$$

To find equation solutions, solve x-6=0 and -x-5=0.

$$x=6 x=-5$$

Substitute 6 for x in the equation \sqrt{31-x}=x-1.

$$\sqrt{31-6}=6-1$$

Simplify. The value x=6 satisfies the equation.

$$5=5$$

Substitute -5 for x in the equation \sqrt{31-x}=x-1.

$$\sqrt{31-\left(-5\right)}=-5-1$$

Simplify. The value x=-5 does not satisfy the equation because the left and the right hand side have opposite signs.

$$6=-6$$

Equation \sqrt{31-x}=x-1 has a unique solution.

$$x=6$$