Solve sqrt{9x+19}=x+3

$$\sqrt { 9 x + 19 } = x + 3$$

Solve for x

$$x=5$$

$$x=-2$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{9x+19}\right)^{2}=\left(x+3\right)^{2}$$

Calculate \sqrt{9x+19} to the power of 2 and get 9x+19.

$$9x+19=\left(x+3\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

$$9x+19=x^{2}+6x+9$$

Subtract x^{2} from both sides.

$$9x+19-x^{2}=6x+9$$

Subtract 6x from both sides.

$$9x+19-x^{2}-6x=9$$

Combine 9x and -6x to get 3x.

$$3x+19-x^{2}=9$$

Subtract 9 from both sides.

$$3x+19-x^{2}-9=0$$

Subtract 9 from 19 to get 10.

$$3x+10-x^{2}=0$$

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

$$-x^{2}+3x+10=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+10. To find a and b, set up a system to be solved.

$$a+b=3 ab=-10=-10$$

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 -10.

$$-1,10 -2,5$$

Calculate the sum for each pair.

$$-1+10=9 -2+5=3$$

The solution is the pair that gives sum 3.

$$a=5 b=-2$$

Rewrite -x^{2}+3x+10 as \left(-x^{2}+5x\right)+\left(-2x+10\right).

$$\left(-x^{2}+5x\right)+\left(-2x+10\right)$$

Factor out -x in the first and -2 in the second group.

$$-x\left(x-5\right)-2\left(x-5\right)$$

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

$$\left(x-5\right)\left(-x-2\right)$$

To find equation solutions, solve x-5=0 and -x-2=0.

$$x=5 x=-2$$

Substitute 5 for x in the equation \sqrt{9x+19}=x+3.

$$\sqrt{9\times 5+19}=5+3$$

Simplify. The value x=5 satisfies the equation.

$$8=8$$

Substitute -2 for x in the equation \sqrt{9x+19}=x+3.

$$\sqrt{9\left(-2\right)+19}=-2+3$$

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

$$1=1$$

List all solutions of \sqrt{9x+19}=x+3.

$$x=5 x=-2$$