Solve 5x+2/5x-2=3x-1/5x+2

$$\frac { 5 x + 2 } { 5 x - 2 } = \frac { 3 x - 1 } { 5 x + 2 }$$

Solve for x

$$x=\frac{\sqrt{881}-31}{20}\approx -0.065917792$$

$$x=\frac{-\sqrt{881}-31}{20}\approx -3.034082208$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -\frac{2}{5},\frac{2}{5} since division by zero is not defined. Multiply both sides of the equation by \left(5x-2\right)\left(5x+2\right), the least common multiple of 5x-2,5x+2.

$$\left(5x+2\right)\left(5x+2\right)=\left(5x-2\right)\left(3x-1\right)$$

Multiply 5x+2 and 5x+2 to get \left(5x+2\right)^{2}.

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

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

$$25x^{2}+20x+4=\left(5x-2\right)\left(3x-1\right)$$

Use the distributive property to multiply 5x-2 by 3x-1 and combine like terms.

$$25x^{2}+20x+4=15x^{2}-11x+2$$

Subtract 15x^{2} from both sides.

$$25x^{2}+20x+4-15x^{2}=-11x+2$$

Combine 25x^{2} and -15x^{2} to get 10x^{2}.

$$10x^{2}+20x+4=-11x+2$$

Add 11x to both sides.

$$10x^{2}+20x+4+11x=2$$

Combine 20x and 11x to get 31x.

$$10x^{2}+31x+4=2$$

Subtract 2 from both sides.

$$10x^{2}+31x+4-2=0$$

Subtract 2 from 4 to get 2.

$$10x^{2}+31x+2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 10 for a, 31 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-31±\sqrt{31^{2}-4\times 10\times 2}}{2\times 10}$$

Square 31.

$$x=\frac{-31±\sqrt{961-4\times 10\times 2}}{2\times 10}$$

Multiply -4 times 10.

$$x=\frac{-31±\sqrt{961-40\times 2}}{2\times 10}$$

Multiply -40 times 2.

$$x=\frac{-31±\sqrt{961-80}}{2\times 10}$$

Add 961 to -80.

$$x=\frac{-31±\sqrt{881}}{2\times 10}$$

Multiply 2 times 10.

$$x=\frac{-31±\sqrt{881}}{20}$$

Now solve the equation x=\frac{-31±\sqrt{881}}{20} when ± is plus. Add -31 to \sqrt{881}.

$$x=\frac{\sqrt{881}-31}{20}$$

Now solve the equation x=\frac{-31±\sqrt{881}}{20} when ± is minus. Subtract \sqrt{881} from -31.

$$x=\frac{-\sqrt{881}-31}{20}$$

The equation is now solved.

$$x=\frac{\sqrt{881}-31}{20} x=\frac{-\sqrt{881}-31}{20}$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -\frac{2}{5},\frac{2}{5} since division by zero is not defined. Multiply both sides of the equation by \left(5x-2\right)\left(5x+2\right), the least common multiple of 5x-2,5x+2.

$$\left(5x+2\right)\left(5x+2\right)=\left(5x-2\right)\left(3x-1\right)$$

Multiply 5x+2 and 5x+2 to get \left(5x+2\right)^{2}.

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

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

$$25x^{2}+20x+4=\left(5x-2\right)\left(3x-1\right)$$

Use the distributive property to multiply 5x-2 by 3x-1 and combine like terms.

$$25x^{2}+20x+4=15x^{2}-11x+2$$

Subtract 15x^{2} from both sides.

$$25x^{2}+20x+4-15x^{2}=-11x+2$$

Combine 25x^{2} and -15x^{2} to get 10x^{2}.

$$10x^{2}+20x+4=-11x+2$$

Add 11x to both sides.

$$10x^{2}+20x+4+11x=2$$

Combine 20x and 11x to get 31x.

$$10x^{2}+31x+4=2$$

Subtract 4 from both sides.

$$10x^{2}+31x=2-4$$

Subtract 4 from 2 to get -2.

$$10x^{2}+31x=-2$$

Divide both sides by 10.

$$\frac{10x^{2}+31x}{10}=-\frac{2}{10}$$

Dividing by 10 undoes the multiplication by 10.

$$x^{2}+\frac{31}{10}x=-\frac{2}{10}$$

Reduce the fraction \frac{-2}{10} to lowest terms by extracting and canceling out 2.

$$x^{2}+\frac{31}{10}x=-\frac{1}{5}$$

Divide \frac{31}{10}, the coefficient of the x term, by 2 to get \frac{31}{20}. Then add the square of \frac{31}{20} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{31}{10}x+\left(\frac{31}{20}\right)^{2}=-\frac{1}{5}+\left(\frac{31}{20}\right)^{2}$$

Square \frac{31}{20} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{31}{10}x+\frac{961}{400}=-\frac{1}{5}+\frac{961}{400}$$

Add -\frac{1}{5} to \frac{961}{400} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{31}{10}x+\frac{961}{400}=\frac{881}{400}$$

Factor x^{2}+\frac{31}{10}x+\frac{961}{400}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{31}{20}\right)^{2}=\frac{881}{400}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{31}{20}\right)^{2}}=\sqrt{\frac{881}{400}}$$

Simplify.

$$x+\frac{31}{20}=\frac{\sqrt{881}}{20} x+\frac{31}{20}=-\frac{\sqrt{881}}{20}$$

Subtract \frac{31}{20} from both sides of the equation.

$$x=\frac{\sqrt{881}-31}{20} x=\frac{-\sqrt{881}-31}{20}$$