Solve 1+q^2/(1+q)^2=117/225=

$$\frac { 1 + q ^ { 2 } } { ( 1 + q ) ^ { 2 } } = \frac { 117 } { 225 } =$$

Solve for q

$$q=\frac{2}{3}\approx 0.666666667$$

$$q = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

Steps Using the Quadratic Formula

Variable q cannot be equal to -1 since division by zero is not defined. Multiply both sides of the equation by 225\left(q+1\right)^{2}, the least common multiple of \left(1+q\right)^{2},225.

$$225\left(1+q^{2}\right)=\left(q+1\right)^{2}\times 117$$

Use the distributive property to multiply 225 by 1+q^{2}.

$$225+225q^{2}=\left(q+1\right)^{2}\times 117$$

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

$$225+225q^{2}=\left(q^{2}+2q+1\right)\times 117$$

Use the distributive property to multiply q^{2}+2q+1 by 117.

$$225+225q^{2}=117q^{2}+234q+117$$

Subtract 117q^{2} from both sides.

$$225+225q^{2}-117q^{2}=234q+117$$

Combine 225q^{2} and -117q^{2} to get 108q^{2}.

$$225+108q^{2}=234q+117$$

Subtract 234q from both sides.

$$225+108q^{2}-234q=117$$

Subtract 117 from both sides.

$$225+108q^{2}-234q-117=0$$

Subtract 117 from 225 to get 108.

$$108+108q^{2}-234q=0$$

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.

$$108q^{2}-234q+108=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 108 for a, -234 for b, and 108 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$q=\frac{-\left(-234\right)±\sqrt{\left(-234\right)^{2}-4\times 108\times 108}}{2\times 108}$$

Square -234.

$$q=\frac{-\left(-234\right)±\sqrt{54756-4\times 108\times 108}}{2\times 108}$$

Multiply -4 times 108.

$$q=\frac{-\left(-234\right)±\sqrt{54756-432\times 108}}{2\times 108}$$

Multiply -432 times 108.

$$q=\frac{-\left(-234\right)±\sqrt{54756-46656}}{2\times 108}$$

Add 54756 to -46656.

$$q=\frac{-\left(-234\right)±\sqrt{8100}}{2\times 108}$$

Take the square root of 8100.

$$q=\frac{-\left(-234\right)±90}{2\times 108}$$

The opposite of -234 is 234.

$$q=\frac{234±90}{2\times 108}$$

Multiply 2 times 108.

$$q=\frac{234±90}{216}$$

Now solve the equation q=\frac{234±90}{216} when ± is plus. Add 234 to 90.

$$q=\frac{324}{216}$$

Reduce the fraction \frac{324}{216} to lowest terms by extracting and canceling out 108.

$$q=\frac{3}{2}$$

Now solve the equation q=\frac{234±90}{216} when ± is minus. Subtract 90 from 234.

$$q=\frac{144}{216}$$

Reduce the fraction \frac{144}{216} to lowest terms by extracting and canceling out 72.

$$q=\frac{2}{3}$$

The equation is now solved.

$$q=\frac{3}{2} q=\frac{2}{3}$$

Steps for Completing the Square

Variable q cannot be equal to -1 since division by zero is not defined. Multiply both sides of the equation by 225\left(q+1\right)^{2}, the least common multiple of \left(1+q\right)^{2},225.

$$225\left(1+q^{2}\right)=\left(q+1\right)^{2}\times 117$$

Use the distributive property to multiply 225 by 1+q^{2}.

$$225+225q^{2}=\left(q+1\right)^{2}\times 117$$

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

$$225+225q^{2}=\left(q^{2}+2q+1\right)\times 117$$

Use the distributive property to multiply q^{2}+2q+1 by 117.

$$225+225q^{2}=117q^{2}+234q+117$$

Subtract 117q^{2} from both sides.

$$225+225q^{2}-117q^{2}=234q+117$$

Combine 225q^{2} and -117q^{2} to get 108q^{2}.

$$225+108q^{2}=234q+117$$

Subtract 234q from both sides.

$$225+108q^{2}-234q=117$$

Subtract 225 from both sides.

$$108q^{2}-234q=117-225$$

Subtract 225 from 117 to get -108.

$$108q^{2}-234q=-108$$

Divide both sides by 108.

$$\frac{108q^{2}-234q}{108}=-\frac{108}{108}$$

Dividing by 108 undoes the multiplication by 108.

$$q^{2}+\left(-\frac{234}{108}\right)q=-\frac{108}{108}$$

Reduce the fraction \frac{-234}{108} to lowest terms by extracting and canceling out 18.

$$q^{2}-\frac{13}{6}q=-\frac{108}{108}$$

Divide -108 by 108.

$$q^{2}-\frac{13}{6}q=-1$$

Divide -\frac{13}{6}, the coefficient of the x term, by 2 to get -\frac{13}{12}. Then add the square of -\frac{13}{12} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$q^{2}-\frac{13}{6}q+\left(-\frac{13}{12}\right)^{2}=-1+\left(-\frac{13}{12}\right)^{2}$$

Square -\frac{13}{12} by squaring both the numerator and the denominator of the fraction.

$$q^{2}-\frac{13}{6}q+\frac{169}{144}=-1+\frac{169}{144}$$

Add -1 to \frac{169}{144}.

$$q^{2}-\frac{13}{6}q+\frac{169}{144}=\frac{25}{144}$$

Factor q^{2}-\frac{13}{6}q+\frac{169}{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}.

$$\left(q-\frac{13}{12}\right)^{2}=\frac{25}{144}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(q-\frac{13}{12}\right)^{2}}=\sqrt{\frac{25}{144}}$$

Simplify.

$$q-\frac{13}{12}=\frac{5}{12} q-\frac{13}{12}=-\frac{5}{12}$$

Add \frac{13}{12} to both sides of the equation.

$$q=\frac{3}{2} q=\frac{2}{3}$$