Solve sqrt{3x+1}+2=sqrt{8x-4}

$$\sqrt{ 3x+1 } +2= \sqrt{ 8x-4 }$$

Solve for x

$$x=5$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{3x+1}+2\right)^{2}=\left(\sqrt{8x-4}\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{3x+1}+2\right)^{2}.

$$\left(\sqrt{3x+1}\right)^{2}+4\sqrt{3x+1}+4=\left(\sqrt{8x-4}\right)^{2}$$

Calculate \sqrt{3x+1} to the power of 2 and get 3x+1.

$$3x+1+4\sqrt{3x+1}+4=\left(\sqrt{8x-4}\right)^{2}$$

Add 1 and 4 to get 5.

$$3x+5+4\sqrt{3x+1}=\left(\sqrt{8x-4}\right)^{2}$$

Calculate \sqrt{8x-4} to the power of 2 and get 8x-4.

$$3x+5+4\sqrt{3x+1}=8x-4$$

Subtract 3x+5 from both sides of the equation.

$$4\sqrt{3x+1}=8x-4-\left(3x+5\right)$$

To find the opposite of 3x+5, find the opposite of each term.

$$4\sqrt{3x+1}=8x-4-3x-5$$

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

$$4\sqrt{3x+1}=5x-4-5$$

Subtract 5 from -4 to get -9.

$$4\sqrt{3x+1}=5x-9$$

Square both sides of the equation.

$$\left(4\sqrt{3x+1}\right)^{2}=\left(5x-9\right)^{2}$$

Expand \left(4\sqrt{3x+1}\right)^{2}.

$$4^{2}\left(\sqrt{3x+1}\right)^{2}=\left(5x-9\right)^{2}$$

Calculate 4 to the power of 2 and get 16.

$$16\left(\sqrt{3x+1}\right)^{2}=\left(5x-9\right)^{2}$$

Calculate \sqrt{3x+1} to the power of 2 and get 3x+1.

$$16\left(3x+1\right)=\left(5x-9\right)^{2}$$

Use the distributive property to multiply 16 by 3x+1.

$$48x+16=\left(5x-9\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(5x-9\right)^{2}.

$$48x+16=25x^{2}-90x+81$$

Subtract 25x^{2} from both sides.

$$48x+16-25x^{2}=-90x+81$$

Add 90x to both sides.

$$48x+16-25x^{2}+90x=81$$

Combine 48x and 90x to get 138x.

$$138x+16-25x^{2}=81$$

Subtract 81 from both sides.

$$138x+16-25x^{2}-81=0$$

Subtract 81 from 16 to get -65.

$$138x-65-25x^{2}=0$$

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.

$$-25x^{2}+138x-65=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -25 for a, 138 for b, and -65 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-138±\sqrt{138^{2}-4\left(-25\right)\left(-65\right)}}{2\left(-25\right)}$$

Square 138.

$$x=\frac{-138±\sqrt{19044-4\left(-25\right)\left(-65\right)}}{2\left(-25\right)}$$

Multiply -4 times -25.

$$x=\frac{-138±\sqrt{19044+100\left(-65\right)}}{2\left(-25\right)}$$

Multiply 100 times -65.

$$x=\frac{-138±\sqrt{19044-6500}}{2\left(-25\right)}$$

Add 19044 to -6500.

$$x=\frac{-138±\sqrt{12544}}{2\left(-25\right)}$$

Take the square root of 12544.

$$x=\frac{-138±112}{2\left(-25\right)}$$

Multiply 2 times -25.

$$x=\frac{-138±112}{-50}$$

Now solve the equation x=\frac{-138±112}{-50} when ± is plus. Add -138 to 112.

$$x=-\frac{26}{-50}$$

Reduce the fraction \frac{-26}{-50} to lowest terms by extracting and canceling out 2.

$$x=\frac{13}{25}$$

Now solve the equation x=\frac{-138±112}{-50} when ± is minus. Subtract 112 from -138.

$$x=-\frac{250}{-50}$$

Divide -250 by -50.

$$x=5$$

The equation is now solved.

$$x=\frac{13}{25} x=5$$

Substitute \frac{13}{25} for x in the equation \sqrt{3x+1}+2=\sqrt{8x-4}.

$$\sqrt{3\times \frac{13}{25}+1}+2=\sqrt{8\times \frac{13}{25}-4}$$

Simplify. The value x=\frac{13}{25} does not satisfy the equation.

$$\frac{18}{5}=\frac{2}{5}$$

Substitute 5 for x in the equation \sqrt{3x+1}+2=\sqrt{8x-4}.

$$\sqrt{3\times 5+1}+2=\sqrt{8\times 5-4}$$

Simplify. The value x=5 satisfies the equation.

$$6=6$$

Equation \sqrt{3x+1}+2=\sqrt{8x-4} has a unique solution.

$$x=5$$