Solve sqrt{x-3}-x=-5

$$\sqrt{ x-3 } -x=-5$$

Solve for x

$$x=7$$

Solution Steps

Subtract -x from both sides of the equation.

$$\sqrt{x-3}=-5+x$$

Square both sides of the equation.

$$\left(\sqrt{x-3}\right)^{2}=\left(-5+x\right)^{2}$$

Calculate \sqrt{x-3} to the power of 2 and get x-3.

$$x-3=\left(-5+x\right)^{2}$$

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

$$x-3=25-10x+x^{2}$$

Subtract 25 from both sides.

$$x-3-25=-10x+x^{2}$$

Subtract 25 from -3 to get -28.

$$x-28=-10x+x^{2}$$

Add 10x to both sides.

$$x-28+10x=x^{2}$$

Combine x and 10x to get 11x.

$$11x-28=x^{2}$$

Subtract x^{2} from both sides.

$$11x-28-x^{2}=0$$

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

$$-x^{2}+11x-28=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-28. To find a and b, set up a system to be solved.

$$a+b=11 ab=-\left(-28\right)=28$$

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 28.

$$1,28 2,14 4,7$$

Calculate the sum for each pair.

$$1+28=29 2+14=16 4+7=11$$

The solution is the pair that gives sum 11.

$$a=7 b=4$$

Rewrite -x^{2}+11x-28 as \left(-x^{2}+7x\right)+\left(4x-28\right).

$$\left(-x^{2}+7x\right)+\left(4x-28\right)$$

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

$$-x\left(x-7\right)+4\left(x-7\right)$$

Factor out common term x-7 by using distributive property.

$$\left(x-7\right)\left(-x+4\right)$$

To find equation solutions, solve x-7=0 and -x+4=0.

$$x=7 x=4$$

Substitute 7 for x in the equation \sqrt{x-3}-x=-5.

$$\sqrt{7-3}-7=-5$$

Simplify. The value x=7 satisfies the equation.

$$-5=-5$$

Substitute 4 for x in the equation \sqrt{x-3}-x=-5.

$$\sqrt{4-3}-4=-5$$

Simplify. The value x=4 does not satisfy the equation.

$$-3=-5$$

Equation \sqrt{x-3}=x-5 has a unique solution.

$$x=7$$