Solve 3x+sqrt{6x+4}=38

$$3x+ \sqrt{ 6x+4 } = 38$$

Solve for x

$$x=10$$

Solution Steps

Subtract 3x from both sides of the equation.

$$\sqrt{6x+4}=38-3x$$

Square both sides of the equation.

$$\left(\sqrt{6x+4}\right)^{2}=\left(38-3x\right)^{2}$$

Calculate \sqrt{6x+4} to the power of 2 and get 6x+4.

$$6x+4=\left(38-3x\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(38-3x\right)^{2}.

$$6x+4=1444-228x+9x^{2}$$

Subtract 1444 from both sides.

$$6x+4-1444=-228x+9x^{2}$$

Subtract 1444 from 4 to get -1440.

$$6x-1440=-228x+9x^{2}$$

Add 228x to both sides.

$$6x-1440+228x=9x^{2}$$

Combine 6x and 228x to get 234x.

$$234x-1440=9x^{2}$$

Subtract 9x^{2} from both sides.

$$234x-1440-9x^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-9x^{2}+234x-1440=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -9 for a, 234 for b, and -1440 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-234±\sqrt{234^{2}-4\left(-9\right)\left(-1440\right)}}{2\left(-9\right)}$$

Square 234.

$$x=\frac{-234±\sqrt{54756-4\left(-9\right)\left(-1440\right)}}{2\left(-9\right)}$$

Multiply -4 times -9.

$$x=\frac{-234±\sqrt{54756+36\left(-1440\right)}}{2\left(-9\right)}$$

Multiply 36 times -1440.

$$x=\frac{-234±\sqrt{54756-51840}}{2\left(-9\right)}$$

Add 54756 to -51840.

$$x=\frac{-234±\sqrt{2916}}{2\left(-9\right)}$$

Take the square root of 2916.

$$x=\frac{-234±54}{2\left(-9\right)}$$

Multiply 2 times -9.

$$x=\frac{-234±54}{-18}$$

Now solve the equation x=\frac{-234±54}{-18} when ± is plus. Add -234 to 54.

$$x=-\frac{180}{-18}$$

Divide -180 by -18.

$$x=10$$

Now solve the equation x=\frac{-234±54}{-18} when ± is minus. Subtract 54 from -234.

$$x=-\frac{288}{-18}$$

Divide -288 by -18.

$$x=16$$

The equation is now solved.

$$x=10 x=16$$

Substitute 10 for x in the equation 3x+\sqrt{6x+4}=38.

$$3\times 10+\sqrt{6\times 10+4}=38$$

Simplify. The value x=10 satisfies the equation.

$$38=38$$

Substitute 16 for x in the equation 3x+\sqrt{6x+4}=38.

$$3\times 16+\sqrt{6\times 16+4}=38$$

Simplify. The value x=16 does not satisfy the equation.

$$58=38$$

Equation \sqrt{6x+4}=38-3x has a unique solution.

$$x=10$$