Solve 1/4-x^2+2=-1/4(x-2)

$$\frac { 1 } { 4 - x ^ { 2 } } + 2 = - \frac { 1 } { 4 ( x - 2 ) }$$

Solve for x

$$x = -\frac{17}{8} = -2\frac{1}{8} = -2.125$$

Steps Using Factoring By Grouping

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

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

Multiply 4 and 2 to get 8.

$$-4+8\left(x-2\right)\left(x+2\right)=-\left(x+2\right)$$

Use the distributive property to multiply 8 by x-2.

$$-4+\left(8x-16\right)\left(x+2\right)=-\left(x+2\right)$$

Use the distributive property to multiply 8x-16 by x+2 and combine like terms.

$$-4+8x^{2}-32=-\left(x+2\right)$$

Subtract 32 from -4 to get -36.

$$-36+8x^{2}=-\left(x+2\right)$$

To find the opposite of x+2, find the opposite of each term.

$$-36+8x^{2}=-x-2$$

Add x to both sides.

$$-36+8x^{2}+x=-2$$

Add 2 to both sides.

$$-36+8x^{2}+x+2=0$$

Add -36 and 2 to get -34.

$$-34+8x^{2}+x=0$$

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

$$8x^{2}+x-34=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 8x^{2}+ax+bx-34. To find a and b, set up a system to be solved.

$$a+b=1 ab=8\left(-34\right)=-272$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -272.

$$-1,272 -2,136 -4,68 -8,34 -16,17$$

Calculate the sum for each pair.

$$-1+272=271 -2+136=134 -4+68=64 -8+34=26 -16+17=1$$

The solution is the pair that gives sum 1.

$$a=-16 b=17$$

Rewrite 8x^{2}+x-34 as \left(8x^{2}-16x\right)+\left(17x-34\right).

$$\left(8x^{2}-16x\right)+\left(17x-34\right)$$

Factor out 8x in the first and 17 in the second group.

$$8x\left(x-2\right)+17\left(x-2\right)$$

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

$$\left(x-2\right)\left(8x+17\right)$$

To find equation solutions, solve x-2=0 and 8x+17=0.

$$x=2 x=-\frac{17}{8}$$

Variable x cannot be equal to 2.

$$x=-\frac{17}{8}$$

Steps Using the Quadratic Formula

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

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

Multiply 4 and 2 to get 8.

$$-4+8\left(x-2\right)\left(x+2\right)=-\left(x+2\right)$$

Use the distributive property to multiply 8 by x-2.

$$-4+\left(8x-16\right)\left(x+2\right)=-\left(x+2\right)$$

Use the distributive property to multiply 8x-16 by x+2 and combine like terms.

$$-4+8x^{2}-32=-\left(x+2\right)$$

Subtract 32 from -4 to get -36.

$$-36+8x^{2}=-\left(x+2\right)$$

To find the opposite of x+2, find the opposite of each term.

$$-36+8x^{2}=-x-2$$

Add x to both sides.

$$-36+8x^{2}+x=-2$$

Add 2 to both sides.

$$-36+8x^{2}+x+2=0$$

Add -36 and 2 to get -34.

$$-34+8x^{2}+x=0$$

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.

$$8x^{2}+x-34=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 8 for a, 1 for b, and -34 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-1±\sqrt{1^{2}-4\times 8\left(-34\right)}}{2\times 8}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\times 8\left(-34\right)}}{2\times 8}$$

Multiply -4 times 8.

$$x=\frac{-1±\sqrt{1-32\left(-34\right)}}{2\times 8}$$

Multiply -32 times -34.

$$x=\frac{-1±\sqrt{1+1088}}{2\times 8}$$

Add 1 to 1088.

$$x=\frac{-1±\sqrt{1089}}{2\times 8}$$

Take the square root of 1089.

$$x=\frac{-1±33}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{-1±33}{16}$$

Now solve the equation x=\frac{-1±33}{16} when ± is plus. Add -1 to 33.

$$x=\frac{32}{16}$$

Divide 32 by 16.

$$x=2$$

Now solve the equation x=\frac{-1±33}{16} when ± is minus. Subtract 33 from -1.

$$x=-\frac{34}{16}$$

Reduce the fraction \frac{-34}{16} to lowest terms by extracting and canceling out 2.

$$x=-\frac{17}{8}$$

The equation is now solved.

$$x=2 x=-\frac{17}{8}$$

Variable x cannot be equal to 2.

$$x=-\frac{17}{8}$$

Steps for Completing the Square

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

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

Multiply 4 and 2 to get 8.

$$-4+8\left(x-2\right)\left(x+2\right)=-\left(x+2\right)$$

Use the distributive property to multiply 8 by x-2.

$$-4+\left(8x-16\right)\left(x+2\right)=-\left(x+2\right)$$

Use the distributive property to multiply 8x-16 by x+2 and combine like terms.

$$-4+8x^{2}-32=-\left(x+2\right)$$

Subtract 32 from -4 to get -36.

$$-36+8x^{2}=-\left(x+2\right)$$

To find the opposite of x+2, find the opposite of each term.

$$-36+8x^{2}=-x-2$$

Add x to both sides.

$$-36+8x^{2}+x=-2$$

Add 36 to both sides.

$$8x^{2}+x=-2+36$$

Add -2 and 36 to get 34.

$$8x^{2}+x=34$$

Divide both sides by 8.

$$\frac{8x^{2}+x}{8}=\frac{34}{8}$$

Dividing by 8 undoes the multiplication by 8.

$$x^{2}+\frac{1}{8}x=\frac{34}{8}$$

Reduce the fraction \frac{34}{8} to lowest terms by extracting and canceling out 2.

$$x^{2}+\frac{1}{8}x=\frac{17}{4}$$

Divide \frac{1}{8}, the coefficient of the x term, by 2 to get \frac{1}{16}. Then add the square of \frac{1}{16} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{8}x+\left(\frac{1}{16}\right)^{2}=\frac{17}{4}+\left(\frac{1}{16}\right)^{2}$$

Square \frac{1}{16} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{8}x+\frac{1}{256}=\frac{17}{4}+\frac{1}{256}$$

Add \frac{17}{4} to \frac{1}{256} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{1}{8}x+\frac{1}{256}=\frac{1089}{256}$$

Factor x^{2}+\frac{1}{8}x+\frac{1}{256}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{16}\right)^{2}=\frac{1089}{256}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{16}\right)^{2}}=\sqrt{\frac{1089}{256}}$$

Simplify.

$$x+\frac{1}{16}=\frac{33}{16} x+\frac{1}{16}=-\frac{33}{16}$$

Subtract \frac{1}{16} from both sides of the equation.

$$x=2 x=-\frac{17}{8}$$

Variable x cannot be equal to 2.

$$x=-\frac{17}{8}$$