Solve sqrt{3x-2}+4x=28

$$\sqrt { 3 x - 2 } + 4 x = 28$$

Solve for x

$$x=6$$

Solution Steps

Subtract 4x from both sides of the equation.

$$\sqrt{3x-2}=28-4x$$

Square both sides of the equation.

$$\left(\sqrt{3x-2}\right)^{2}=\left(28-4x\right)^{2}$$

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

$$3x-2=\left(28-4x\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(28-4x\right)^{2}.

$$3x-2=784-224x+16x^{2}$$

Subtract 784 from both sides.

$$3x-2-784=-224x+16x^{2}$$

Subtract 784 from -2 to get -786.

$$3x-786=-224x+16x^{2}$$

Add 224x to both sides.

$$3x-786+224x=16x^{2}$$

Combine 3x and 224x to get 227x.

$$227x-786=16x^{2}$$

Subtract 16x^{2} from both sides.

$$227x-786-16x^{2}=0$$

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.

$$-16x^{2}+227x-786=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -16 for a, 227 for b, and -786 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-227±\sqrt{227^{2}-4\left(-16\right)\left(-786\right)}}{2\left(-16\right)}$$

Square 227.

$$x=\frac{-227±\sqrt{51529-4\left(-16\right)\left(-786\right)}}{2\left(-16\right)}$$

Multiply -4 times -16.

$$x=\frac{-227±\sqrt{51529+64\left(-786\right)}}{2\left(-16\right)}$$

Multiply 64 times -786.

$$x=\frac{-227±\sqrt{51529-50304}}{2\left(-16\right)}$$

Add 51529 to -50304.

$$x=\frac{-227±\sqrt{1225}}{2\left(-16\right)}$$

Take the square root of 1225.

$$x=\frac{-227±35}{2\left(-16\right)}$$

Multiply 2 times -16.

$$x=\frac{-227±35}{-32}$$

Now solve the equation x=\frac{-227±35}{-32} when ± is plus. Add -227 to 35.

$$x=-\frac{192}{-32}$$

Divide -192 by -32.

$$x=6$$

Now solve the equation x=\frac{-227±35}{-32} when ± is minus. Subtract 35 from -227.

$$x=-\frac{262}{-32}$$

Reduce the fraction \frac{-262}{-32} to lowest terms by extracting and canceling out 2.

$$x=\frac{131}{16}$$

The equation is now solved.

$$x=6 x=\frac{131}{16}$$

Substitute 6 for x in the equation \sqrt{3x-2}+4x=28.

$$\sqrt{3\times 6-2}+4\times 6=28$$

Simplify. The value x=6 satisfies the equation.

$$28=28$$

Substitute \frac{131}{16} for x in the equation \sqrt{3x-2}+4x=28.

$$\sqrt{3\times \frac{131}{16}-2}+4\times \frac{131}{16}=28$$

Simplify. The value x=\frac{131}{16} does not satisfy the equation.

$$\frac{75}{2}=28$$

Equation \sqrt{3x-2}=28-4x has a unique solution.

$$x=6$$