Solve sqrt{3x+10}=x+4

$$\sqrt{ 3x+10 } =x+4$$

Solve for x

$$x=-2$$

$$x=-3$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{3x+10}\right)^{2}=\left(x+4\right)^{2}$$

Calculate \sqrt{3x+10} to the power of 2 and get 3x+10.

$$3x+10=\left(x+4\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+4\right)^{2}.

$$3x+10=x^{2}+8x+16$$

Subtract x^{2} from both sides.

$$3x+10-x^{2}=8x+16$$

Subtract 8x from both sides.

$$3x+10-x^{2}-8x=16$$

Combine 3x and -8x to get -5x.

$$-5x+10-x^{2}=16$$

Subtract 16 from both sides.

$$-5x+10-x^{2}-16=0$$

Subtract 16 from 10 to get -6.

$$-5x-6-x^{2}=0$$

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

$$-x^{2}-5x-6=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx-6. To find a and b, set up a system to be solved.

$$a+b=-5 ab=-\left(-6\right)=6$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 6.

$$-1,-6 -2,-3$$

Calculate the sum for each pair.

$$-1-6=-7 -2-3=-5$$

The solution is the pair that gives sum -5.

$$a=-2 b=-3$$

Rewrite -x^{2}-5x-6 as \left(-x^{2}-2x\right)+\left(-3x-6\right).

$$\left(-x^{2}-2x\right)+\left(-3x-6\right)$$

Factor out x in the first and 3 in the second group.

$$x\left(-x-2\right)+3\left(-x-2\right)$$

Factor out common term -x-2 by using distributive property.

$$\left(-x-2\right)\left(x+3\right)$$

To find equation solutions, solve -x-2=0 and x+3=0.

$$x=-2 x=-3$$

Substitute -2 for x in the equation \sqrt{3x+10}=x+4.

$$\sqrt{3\left(-2\right)+10}=-2+4$$

Simplify. The value x=-2 satisfies the equation.

$$2=2$$

Substitute -3 for x in the equation \sqrt{3x+10}=x+4.

$$\sqrt{3\left(-3\right)+10}=-3+4$$

Simplify. The value x=-3 satisfies the equation.

$$1=1$$

List all solutions of \sqrt{3x+10}=x+4.

$$x=-2 x=-3$$