Solve x^2+x/6+9/16-2/3=0

$$x ^ { 2 } + \frac { x } { 6 } + \frac { 9 } { 16 } - \frac { 2 } { 3 } = 0$$

Solve for x

$$x=-\frac{5}{12}\approx -0.416666667$$

$$x=\frac{1}{4}=0.25$$

Steps Using Factoring By Grouping

Multiply both sides of the equation by 48, the least common multiple of 6,16,3.

$$48x^{2}+8x+27-32=0$$

Subtract 32 from 27 to get -5.

$$48x^{2}+8x-5=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 48x^{2}+ax+bx-5. To find a and b, set up a system to be solved.

$$a+b=8 ab=48\left(-5\right)=-240$$

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 -240.

$$-1,240 -2,120 -3,80 -4,60 -5,48 -6,40 -8,30 -10,24 -12,20 -15,16$$

Calculate the sum for each pair.

$$-1+240=239 -2+120=118 -3+80=77 -4+60=56 -5+48=43 -6+40=34 -8+30=22 -10+24=14 -12+20=8 -15+16=1$$

The solution is the pair that gives sum 8.

$$a=-12 b=20$$

Rewrite 48x^{2}+8x-5 as \left(48x^{2}-12x\right)+\left(20x-5\right).

$$\left(48x^{2}-12x\right)+\left(20x-5\right)$$

Factor out 12x in the first and 5 in the second group.

$$12x\left(4x-1\right)+5\left(4x-1\right)$$

Factor out common term 4x-1 by using distributive property.

$$\left(4x-1\right)\left(12x+5\right)$$

To find equation solutions, solve 4x-1=0 and 12x+5=0.

$$x=\frac{1}{4} x=-\frac{5}{12}$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 48, the least common multiple of 6,16,3.

$$48x^{2}+8x+27-32=0$$

Subtract 32 from 27 to get -5.

$$48x^{2}+8x-5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 48 for a, 8 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-8±\sqrt{8^{2}-4\times 48\left(-5\right)}}{2\times 48}$$

Square 8.

$$x=\frac{-8±\sqrt{64-4\times 48\left(-5\right)}}{2\times 48}$$

Multiply -4 times 48.

$$x=\frac{-8±\sqrt{64-192\left(-5\right)}}{2\times 48}$$

Multiply -192 times -5.

$$x=\frac{-8±\sqrt{64+960}}{2\times 48}$$

Add 64 to 960.

$$x=\frac{-8±\sqrt{1024}}{2\times 48}$$

Take the square root of 1024.

$$x=\frac{-8±32}{2\times 48}$$

Multiply 2 times 48.

$$x=\frac{-8±32}{96}$$

Now solve the equation x=\frac{-8±32}{96} when ± is plus. Add -8 to 32.

$$x=\frac{24}{96}$$

Reduce the fraction \frac{24}{96} to lowest terms by extracting and canceling out 24.

$$x=\frac{1}{4}$$

Now solve the equation x=\frac{-8±32}{96} when ± is minus. Subtract 32 from -8.

$$x=-\frac{40}{96}$$

Reduce the fraction \frac{-40}{96} to lowest terms by extracting and canceling out 8.

$$x=-\frac{5}{12}$$

The equation is now solved.

$$x=\frac{1}{4} x=-\frac{5}{12}$$

Steps for Completing the Square

Multiply both sides of the equation by 48, the least common multiple of 6,16,3.

$$48x^{2}+8x+27-32=0$$

Subtract 32 from 27 to get -5.

$$48x^{2}+8x-5=0$$

Add 5 to both sides. Anything plus zero gives itself.

$$48x^{2}+8x=5$$

Divide both sides by 48.

$$\frac{48x^{2}+8x}{48}=\frac{5}{48}$$

Dividing by 48 undoes the multiplication by 48.

$$x^{2}+\frac{8}{48}x=\frac{5}{48}$$

Reduce the fraction \frac{8}{48} to lowest terms by extracting and canceling out 8.

$$x^{2}+\frac{1}{6}x=\frac{5}{48}$$

Divide \frac{1}{6}, the coefficient of the x term, by 2 to get \frac{1}{12}. Then add the square of \frac{1}{12} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{6}x+\left(\frac{1}{12}\right)^{2}=\frac{5}{48}+\left(\frac{1}{12}\right)^{2}$$

Square \frac{1}{12} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{6}x+\frac{1}{144}=\frac{5}{48}+\frac{1}{144}$$

Add \frac{5}{48} to \frac{1}{144} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{1}{6}x+\frac{1}{144}=\frac{1}{9}$$

Factor x^{2}+\frac{1}{6}x+\frac{1}{144}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{12}\right)^{2}=\frac{1}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{12}\right)^{2}}=\sqrt{\frac{1}{9}}$$

Simplify.

$$x+\frac{1}{12}=\frac{1}{3} x+\frac{1}{12}=-\frac{1}{3}$$

Subtract \frac{1}{12} from both sides of the equation.

$$x=\frac{1}{4} x=-\frac{5}{12}$$