Solve (x+3)(6-x)/36-4x^2=-1/4

$$\frac { ( x + 3 ) ( 6 - x ) } { 36 - 4 x ^ { 2 } } = - \frac { 1 } { 4 }$$

Solve for x

$$x = \frac{9}{2} = 4\frac{1}{2} = 4.5$$

Steps Using Factoring By Grouping

Variable x cannot be equal to any of the values -3,3 since division by zero is not defined. Multiply both sides of the equation by 4\left(x-3\right)\left(x+3\right), the least common multiple of 36-4x^{2},4.

$$-\left(x+3\right)\left(6-x\right)=-\left(x-3\right)\left(x+3\right)$$

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

$$\left(-x-3\right)\left(6-x\right)=-\left(x-3\right)\left(x+3\right)$$

Use the distributive property to multiply -x-3 by 6-x and combine like terms.

$$-3x+x^{2}-18=-\left(x-3\right)\left(x+3\right)$$

Use the distributive property to multiply -1 by x-3.

$$-3x+x^{2}-18=\left(-x+3\right)\left(x+3\right)$$

Use the distributive property to multiply -x+3 by x+3 and combine like terms.

$$-3x+x^{2}-18=-x^{2}+9$$

Add x^{2} to both sides.

$$-3x+x^{2}-18+x^{2}=9$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$-3x+2x^{2}-18=9$$

Subtract 9 from both sides.

$$-3x+2x^{2}-18-9=0$$

Subtract 9 from -18 to get -27.

$$-3x+2x^{2}-27=0$$

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

$$2x^{2}-3x-27=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx-27. To find a and b, set up a system to be solved.

$$a+b=-3 ab=2\left(-27\right)=-54$$

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 -54.

$$1,-54 2,-27 3,-18 6,-9$$

Calculate the sum for each pair.

$$1-54=-53 2-27=-25 3-18=-15 6-9=-3$$

The solution is the pair that gives sum -3.

$$a=-9 b=6$$

Rewrite 2x^{2}-3x-27 as \left(2x^{2}-9x\right)+\left(6x-27\right).

$$\left(2x^{2}-9x\right)+\left(6x-27\right)$$

Factor out x in the first and 3 in the second group.

$$x\left(2x-9\right)+3\left(2x-9\right)$$

Factor out common term 2x-9 by using distributive property.

$$\left(2x-9\right)\left(x+3\right)$$

To find equation solutions, solve 2x-9=0 and x+3=0.

$$x=\frac{9}{2} x=-3$$

Variable x cannot be equal to -3.

$$x=\frac{9}{2}$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -3,3 since division by zero is not defined. Multiply both sides of the equation by 4\left(x-3\right)\left(x+3\right), the least common multiple of 36-4x^{2},4.

$$-\left(x+3\right)\left(6-x\right)=-\left(x-3\right)\left(x+3\right)$$

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

$$\left(-x-3\right)\left(6-x\right)=-\left(x-3\right)\left(x+3\right)$$

Use the distributive property to multiply -x-3 by 6-x and combine like terms.

$$-3x+x^{2}-18=-\left(x-3\right)\left(x+3\right)$$

Use the distributive property to multiply -1 by x-3.

$$-3x+x^{2}-18=\left(-x+3\right)\left(x+3\right)$$

Use the distributive property to multiply -x+3 by x+3 and combine like terms.

$$-3x+x^{2}-18=-x^{2}+9$$

Add x^{2} to both sides.

$$-3x+x^{2}-18+x^{2}=9$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$-3x+2x^{2}-18=9$$

Subtract 9 from both sides.

$$-3x+2x^{2}-18-9=0$$

Subtract 9 from -18 to get -27.

$$-3x+2x^{2}-27=0$$

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.

$$2x^{2}-3x-27=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -3 for b, and -27 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 2\left(-27\right)}}{2\times 2}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\times 2\left(-27\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-3\right)±\sqrt{9-8\left(-27\right)}}{2\times 2}$$

Multiply -8 times -27.

$$x=\frac{-\left(-3\right)±\sqrt{9+216}}{2\times 2}$$

Add 9 to 216.

$$x=\frac{-\left(-3\right)±\sqrt{225}}{2\times 2}$$

Take the square root of 225.

$$x=\frac{-\left(-3\right)±15}{2\times 2}$$

The opposite of -3 is 3.

$$x=\frac{3±15}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{3±15}{4}$$

Now solve the equation x=\frac{3±15}{4} when ± is plus. Add 3 to 15.

$$x=\frac{18}{4}$$

Reduce the fraction \frac{18}{4} to lowest terms by extracting and canceling out 2.

$$x=\frac{9}{2}$$

Now solve the equation x=\frac{3±15}{4} when ± is minus. Subtract 15 from 3.

$$x=-\frac{12}{4}$$

Divide -12 by 4.

$$x=-3$$

The equation is now solved.

$$x=\frac{9}{2} x=-3$$

Variable x cannot be equal to -3.

$$x=\frac{9}{2}$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -3,3 since division by zero is not defined. Multiply both sides of the equation by 4\left(x-3\right)\left(x+3\right), the least common multiple of 36-4x^{2},4.

$$-\left(x+3\right)\left(6-x\right)=-\left(x-3\right)\left(x+3\right)$$

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

$$\left(-x-3\right)\left(6-x\right)=-\left(x-3\right)\left(x+3\right)$$

Use the distributive property to multiply -x-3 by 6-x and combine like terms.

$$-3x+x^{2}-18=-\left(x-3\right)\left(x+3\right)$$

Use the distributive property to multiply -1 by x-3.

$$-3x+x^{2}-18=\left(-x+3\right)\left(x+3\right)$$

Use the distributive property to multiply -x+3 by x+3 and combine like terms.

$$-3x+x^{2}-18=-x^{2}+9$$

Add x^{2} to both sides.

$$-3x+x^{2}-18+x^{2}=9$$

Combine x^{2} and x^{2} to get 2x^{2}.

$$-3x+2x^{2}-18=9$$

Add 18 to both sides.

$$-3x+2x^{2}=9+18$$

Add 9 and 18 to get 27.

$$-3x+2x^{2}=27$$

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.

$$2x^{2}-3x=27$$

Divide both sides by 2.

$$\frac{2x^{2}-3x}{2}=\frac{27}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}-\frac{3}{2}x=\frac{27}{2}$$

Divide -\frac{3}{2}, the coefficient of the x term, by 2 to get -\frac{3}{4}. Then add the square of -\frac{3}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{3}{2}x+\left(-\frac{3}{4}\right)^{2}=\frac{27}{2}+\left(-\frac{3}{4}\right)^{2}$$

Square -\frac{3}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=\frac{27}{2}+\frac{9}{16}$$

Add \frac{27}{2} to \frac{9}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=\frac{225}{16}$$

Factor x^{2}-\frac{3}{2}x+\frac{9}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{3}{4}\right)^{2}=\frac{225}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{4}\right)^{2}}=\sqrt{\frac{225}{16}}$$

Simplify.

$$x-\frac{3}{4}=\frac{15}{4} x-\frac{3}{4}=-\frac{15}{4}$$

Add \frac{3}{4} to both sides of the equation.

$$x=\frac{9}{2} x=-3$$

Variable x cannot be equal to -3.

$$x=\frac{9}{2}$$