Solve sqrt{37-x}=x-7

$$\sqrt { 37 - x } = x - 7$$

Solve for x

$$x=12$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{37-x}\right)^{2}=\left(x-7\right)^{2}$$

Calculate \sqrt{37-x} to the power of 2 and get 37-x.

$$37-x=\left(x-7\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-7\right)^{2}.

$$37-x=x^{2}-14x+49$$

Subtract x^{2} from both sides.

$$37-x-x^{2}=-14x+49$$

Add 14x to both sides.

$$37-x-x^{2}+14x=49$$

Combine -x and 14x to get 13x.

$$37+13x-x^{2}=49$$

Subtract 49 from both sides.

$$37+13x-x^{2}-49=0$$

Subtract 49 from 37 to get -12.

$$-12+13x-x^{2}=0$$

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

$$-x^{2}+13x-12=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-12. To find a and b, set up a system to be solved.

$$a+b=13 ab=-\left(-12\right)=12$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 12.

$$1,12 2,6 3,4$$

Calculate the sum for each pair.

$$1+12=13 2+6=8 3+4=7$$

The solution is the pair that gives sum 13.

$$a=12 b=1$$

Rewrite -x^{2}+13x-12 as \left(-x^{2}+12x\right)+\left(x-12\right).

$$\left(-x^{2}+12x\right)+\left(x-12\right)$$

Factor out -x in -x^{2}+12x.

$$-x\left(x-12\right)+x-12$$

Factor out common term x-12 by using distributive property.

$$\left(x-12\right)\left(-x+1\right)$$

To find equation solutions, solve x-12=0 and -x+1=0.

$$x=12 x=1$$

Substitute 12 for x in the equation \sqrt{37-x}=x-7.

$$\sqrt{37-12}=12-7$$

Simplify. The value x=12 satisfies the equation.

$$5=5$$

Substitute 1 for x in the equation \sqrt{37-x}=x-7.

$$\sqrt{37-1}=1-7$$

Simplify. The value x=1 does not satisfy the equation because the left and the right hand side have opposite signs.

$$6=-6$$

Equation \sqrt{37-x}=x-7 has a unique solution.

$$x=12$$