Solve sqrt{6x+19}-4=x

$$\sqrt { 6 x + 19 } - 4 = x$$

Solve for x

$$x=1$$

$$x=-3$$

Solution Steps

Subtract -4 from both sides of the equation.

$$\sqrt{6x+19}=x+4$$

Square both sides of the equation.

$$\left(\sqrt{6x+19}\right)^{2}=\left(x+4\right)^{2}$$

Calculate \sqrt{6x+19} to the power of 2 and get 6x+19.

$$6x+19=\left(x+4\right)^{2}$$

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

$$6x+19=x^{2}+8x+16$$

Subtract x^{2} from both sides.

$$6x+19-x^{2}=8x+16$$

Subtract 8x from both sides.

$$6x+19-x^{2}-8x=16$$

Combine 6x and -8x to get -2x.

$$-2x+19-x^{2}=16$$

Subtract 16 from both sides.

$$-2x+19-x^{2}-16=0$$

Subtract 16 from 19 to get 3.

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

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

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+3. To find a and b, set up a system to be solved.

$$a+b=-2 ab=-3=-3$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. The only such pair is the system solution.

$$a=1 b=-3$$

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

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

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

$$x\left(-x+1\right)+3\left(-x+1\right)$$

Factor out common term -x+1 by using distributive property.

$$\left(-x+1\right)\left(x+3\right)$$

To find equation solutions, solve -x+1=0 and x+3=0.

$$x=1 x=-3$$

Substitute 1 for x in the equation \sqrt{6x+19}-4=x.

$$\sqrt{6\times 1+19}-4=1$$

Simplify. The value x=1 satisfies the equation.

$$1=1$$

Substitute -3 for x in the equation \sqrt{6x+19}-4=x.

$$\sqrt{6\left(-3\right)+19}-4=-3$$

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

$$-3=-3$$

List all solutions of \sqrt{6x+19}=x+4.

$$x=1 x=-3$$