Solve 1-5x/4x^2-7x-2-x/x-2=2/4x+1

$$\frac { 1 - 5 x } { 4 x ^ { 2 } - 7 x - 2 } - \frac { x } { x - 2 } = \frac { 2 } { 4 x + 1 }$$

Solve for x

$$x = -\frac{5}{2} = -2\frac{1}{2} = -2.5$$

$$x=\frac{1}{2}=0.5$$

Steps Using Factoring By Grouping

Variable x cannot be equal to any of the values -\frac{1}{4},2 since division by zero is not defined. Multiply both sides of the equation by \left(x-2\right)\left(4x+1\right), the least common multiple of 4x^{2}-7x-2,x-2,4x+1.

$$1-5x-\left(4x+1\right)x=\left(x-2\right)\times 2$$

Use the distributive property to multiply 4x+1 by x.

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

To find the opposite of 4x^{2}+x, find the opposite of each term.

$$1-5x-4x^{2}-x=\left(x-2\right)\times 2$$

Combine -5x and -x to get -6x.

$$1-6x-4x^{2}=\left(x-2\right)\times 2$$

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

$$1-6x-4x^{2}=2x-4$$

Subtract 2x from both sides.

$$1-6x-4x^{2}-2x=-4$$

Combine -6x and -2x to get -8x.

$$1-8x-4x^{2}=-4$$

Add 4 to both sides.

$$1-8x-4x^{2}+4=0$$

Add 1 and 4 to get 5.

$$5-8x-4x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-4x^{2}-8x+5=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -4x^{2}+ax+bx+5. To find a and b, set up a system to be solved.

$$a+b=-8 ab=-4\times 5=-20$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -20.

$$1,-20 2,-10 4,-5$$

Calculate the sum for each pair.

$$1-20=-19 2-10=-8 4-5=-1$$

The solution is the pair that gives sum -8.

$$a=2 b=-10$$

Rewrite -4x^{2}-8x+5 as \left(-4x^{2}+2x\right)+\left(-10x+5\right).

$$\left(-4x^{2}+2x\right)+\left(-10x+5\right)$$

Factor out 2x in the first and 5 in the second group.

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

Factor out common term -2x+1 by using distributive property.

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

To find equation solutions, solve -2x+1=0 and 2x+5=0.

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

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -\frac{1}{4},2 since division by zero is not defined. Multiply both sides of the equation by \left(x-2\right)\left(4x+1\right), the least common multiple of 4x^{2}-7x-2,x-2,4x+1.

$$1-5x-\left(4x+1\right)x=\left(x-2\right)\times 2$$

Use the distributive property to multiply 4x+1 by x.

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

To find the opposite of 4x^{2}+x, find the opposite of each term.

$$1-5x-4x^{2}-x=\left(x-2\right)\times 2$$

Combine -5x and -x to get -6x.

$$1-6x-4x^{2}=\left(x-2\right)\times 2$$

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

$$1-6x-4x^{2}=2x-4$$

Subtract 2x from both sides.

$$1-6x-4x^{2}-2x=-4$$

Combine -6x and -2x to get -8x.

$$1-8x-4x^{2}=-4$$

Add 4 to both sides.

$$1-8x-4x^{2}+4=0$$

Add 1 and 4 to get 5.

$$5-8x-4x^{2}=0$$

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.

$$-4x^{2}-8x+5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, -8 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\left(-4\right)\times 5}}{2\left(-4\right)}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\left(-4\right)\times 5}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{-\left(-8\right)±\sqrt{64+16\times 5}}{2\left(-4\right)}$$

Multiply 16 times 5.

$$x=\frac{-\left(-8\right)±\sqrt{64+80}}{2\left(-4\right)}$$

Add 64 to 80.

$$x=\frac{-\left(-8\right)±\sqrt{144}}{2\left(-4\right)}$$

Take the square root of 144.

$$x=\frac{-\left(-8\right)±12}{2\left(-4\right)}$$

The opposite of -8 is 8.

$$x=\frac{8±12}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{8±12}{-8}$$

Now solve the equation x=\frac{8±12}{-8} when ± is plus. Add 8 to 12.

$$x=\frac{20}{-8}$$

Reduce the fraction \frac{20}{-8} to lowest terms by extracting and canceling out 4.

$$x=-\frac{5}{2}$$

Now solve the equation x=\frac{8±12}{-8} when ± is minus. Subtract 12 from 8.

$$x=-\frac{4}{-8}$$

Reduce the fraction \frac{-4}{-8} to lowest terms by extracting and canceling out 4.

$$x=\frac{1}{2}$$

The equation is now solved.

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

Steps for Completing the Square

Variable x cannot be equal to any of the values -\frac{1}{4},2 since division by zero is not defined. Multiply both sides of the equation by \left(x-2\right)\left(4x+1\right), the least common multiple of 4x^{2}-7x-2,x-2,4x+1.

$$1-5x-\left(4x+1\right)x=\left(x-2\right)\times 2$$

Use the distributive property to multiply 4x+1 by x.

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

To find the opposite of 4x^{2}+x, find the opposite of each term.

$$1-5x-4x^{2}-x=\left(x-2\right)\times 2$$

Combine -5x and -x to get -6x.

$$1-6x-4x^{2}=\left(x-2\right)\times 2$$

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

$$1-6x-4x^{2}=2x-4$$

Subtract 2x from both sides.

$$1-6x-4x^{2}-2x=-4$$

Combine -6x and -2x to get -8x.

$$1-8x-4x^{2}=-4$$

Subtract 1 from both sides.

$$-8x-4x^{2}=-4-1$$

Subtract 1 from -4 to get -5.

$$-8x-4x^{2}=-5$$

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.

$$-4x^{2}-8x=-5$$

Divide both sides by -4.

$$\frac{-4x^{2}-8x}{-4}=-\frac{5}{-4}$$

Dividing by -4 undoes the multiplication by -4.

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

Divide -8 by -4.

$$x^{2}+2x=-\frac{5}{-4}$$

Divide -5 by -4.

$$x^{2}+2x=\frac{5}{4}$$

Divide 2, the coefficient of the x term, by 2 to get 1. Then add the square of 1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+2x+1^{2}=\frac{5}{4}+1^{2}$$

Square 1.

$$x^{2}+2x+1=\frac{5}{4}+1$$

Add \frac{5}{4} to 1.

$$x^{2}+2x+1=\frac{9}{4}$$

Factor x^{2}+2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+1\right)^{2}=\frac{9}{4}$$

Take the square root of both sides of the equation.

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

Simplify.

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

Subtract 1 from both sides of the equation.

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