Solve 5/(2x-1)^2-10/2x-1=1

$$\frac { 5 } { ( 2 x - 1 ) ^ { 2 } } - \frac { 10 } { 2 x - 1 } = 1$$

Solve for x

$$x=-\frac{\sqrt{30}}{2}-2\approx -4.738612788$$

$$x=\frac{\sqrt{30}}{2}-2\approx 0.738612788$$

Steps Using the Quadratic Formula

Variable x cannot be equal to \frac{1}{2} since division by zero is not defined. Multiply both sides of the equation by \left(2x-1\right)^{2}, the least common multiple of \left(2x-1\right)^{2},2x-1.

$$5-\left(2x-1\right)\times 10=\left(2x-1\right)^{2}$$

Use the distributive property to multiply 2x-1 by 10.

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

To find the opposite of 20x-10, find the opposite of each term.

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

Add 5 and 10 to get 15.

$$15-20x=\left(2x-1\right)^{2}$$

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

$$15-20x=4x^{2}-4x+1$$

Subtract 4x^{2} from both sides.

$$15-20x-4x^{2}=-4x+1$$

Add 4x to both sides.

$$15-20x-4x^{2}+4x=1$$

Combine -20x and 4x to get -16x.

$$15-16x-4x^{2}=1$$

Subtract 1 from both sides.

$$15-16x-4x^{2}-1=0$$

Subtract 1 from 15 to get 14.

$$14-16x-4x^{2}=0$$

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.

$$-4x^{2}-16x+14=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, -16 for b, and 14 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\left(-4\right)\times 14}}{2\left(-4\right)}$$

Square -16.

$$x=\frac{-\left(-16\right)±\sqrt{256-4\left(-4\right)\times 14}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{-\left(-16\right)±\sqrt{256+16\times 14}}{2\left(-4\right)}$$

Multiply 16 times 14.

$$x=\frac{-\left(-16\right)±\sqrt{256+224}}{2\left(-4\right)}$$

Add 256 to 224.

$$x=\frac{-\left(-16\right)±\sqrt{480}}{2\left(-4\right)}$$

Take the square root of 480.

$$x=\frac{-\left(-16\right)±4\sqrt{30}}{2\left(-4\right)}$$

The opposite of -16 is 16.

$$x=\frac{16±4\sqrt{30}}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{16±4\sqrt{30}}{-8}$$

Now solve the equation x=\frac{16±4\sqrt{30}}{-8} when ± is plus. Add 16 to 4\sqrt{30}.

$$x=\frac{4\sqrt{30}+16}{-8}$$

Divide 16+4\sqrt{30} by -8.

$$x=-\frac{\sqrt{30}}{2}-2$$

Now solve the equation x=\frac{16±4\sqrt{30}}{-8} when ± is minus. Subtract 4\sqrt{30} from 16.

$$x=\frac{16-4\sqrt{30}}{-8}$$

Divide 16-4\sqrt{30} by -8.

$$x=\frac{\sqrt{30}}{2}-2$$

The equation is now solved.

$$x=-\frac{\sqrt{30}}{2}-2 x=\frac{\sqrt{30}}{2}-2$$

Steps for Completing the Square

Variable x cannot be equal to \frac{1}{2} since division by zero is not defined. Multiply both sides of the equation by \left(2x-1\right)^{2}, the least common multiple of \left(2x-1\right)^{2},2x-1.

$$5-\left(2x-1\right)\times 10=\left(2x-1\right)^{2}$$

Use the distributive property to multiply 2x-1 by 10.

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

To find the opposite of 20x-10, find the opposite of each term.

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

Add 5 and 10 to get 15.

$$15-20x=\left(2x-1\right)^{2}$$

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

$$15-20x=4x^{2}-4x+1$$

Subtract 4x^{2} from both sides.

$$15-20x-4x^{2}=-4x+1$$

Add 4x to both sides.

$$15-20x-4x^{2}+4x=1$$

Combine -20x and 4x to get -16x.

$$15-16x-4x^{2}=1$$

Subtract 15 from both sides.

$$-16x-4x^{2}=1-15$$

Subtract 15 from 1 to get -14.

$$-16x-4x^{2}=-14$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-4x^{2}-16x=-14$$

Divide both sides by -4.

$$\frac{-4x^{2}-16x}{-4}=-\frac{14}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$x^{2}+\left(-\frac{16}{-4}\right)x=-\frac{14}{-4}$$

Divide -16 by -4.

$$x^{2}+4x=-\frac{14}{-4}$$

Reduce the fraction \frac{-14}{-4} to lowest terms by extracting and canceling out 2.

$$x^{2}+4x=\frac{7}{2}$$

Divide 4, the coefficient of the x term, by 2 to get 2. Then add the square of 2 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+4x+2^{2}=\frac{7}{2}+2^{2}$$

Square 2.

$$x^{2}+4x+4=\frac{7}{2}+4$$

Add \frac{7}{2} to 4.

$$x^{2}+4x+4=\frac{15}{2}$$

Factor x^{2}+4x+4. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+2\right)^{2}=\frac{15}{2}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+2\right)^{2}}=\sqrt{\frac{15}{2}}$$

Simplify.

$$x+2=\frac{\sqrt{30}}{2} x+2=-\frac{\sqrt{30}}{2}$$

Subtract 2 from both sides of the equation.

$$x=\frac{\sqrt{30}}{2}-2 x=-\frac{\sqrt{30}}{2}-2$$