Solve 1/4+9k^2+12k=25/4+9k^2

$$\frac { 1 } { 4 + 9 k ^ { 2 } + 12 k } = \frac { 25 } { 4 + 9 k ^ { 2 } }$$

Solve for k

$$k=-\frac{1}{2}=-0.5$$

$$k=-\frac{8}{9}\approx -0.888888889$$

Steps Using the Quadratic Formula

Variable k cannot be equal to -\frac{2}{3} since division by zero is not defined. Multiply both sides of the equation by \left(3k+2\right)^{2}\left(9k^{2}+4\right), the least common multiple of 4+9k^{2}+12k,4+9k^{2}.

$$9k^{2}+4=\left(3k+2\right)^{2}\times 25$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3k+2\right)^{2}.

$$9k^{2}+4=\left(9k^{2}+12k+4\right)\times 25$$

Use the distributive property to multiply 9k^{2}+12k+4 by 25.

$$9k^{2}+4=225k^{2}+300k+100$$

Subtract 225k^{2} from both sides.

$$9k^{2}+4-225k^{2}=300k+100$$

Combine 9k^{2} and -225k^{2} to get -216k^{2}.

$$-216k^{2}+4=300k+100$$

Subtract 300k from both sides.

$$-216k^{2}+4-300k=100$$

Subtract 100 from both sides.

$$-216k^{2}+4-300k-100=0$$

Subtract 100 from 4 to get -96.

$$-216k^{2}-96-300k=0$$

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.

$$-216k^{2}-300k-96=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -216 for a, -300 for b, and -96 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$k=\frac{-\left(-300\right)±\sqrt{\left(-300\right)^{2}-4\left(-216\right)\left(-96\right)}}{2\left(-216\right)}$$

Square -300.

$$k=\frac{-\left(-300\right)±\sqrt{90000-4\left(-216\right)\left(-96\right)}}{2\left(-216\right)}$$

Multiply -4 times -216.

$$k=\frac{-\left(-300\right)±\sqrt{90000+864\left(-96\right)}}{2\left(-216\right)}$$

Multiply 864 times -96.

$$k=\frac{-\left(-300\right)±\sqrt{90000-82944}}{2\left(-216\right)}$$

Add 90000 to -82944.

$$k=\frac{-\left(-300\right)±\sqrt{7056}}{2\left(-216\right)}$$

Take the square root of 7056.

$$k=\frac{-\left(-300\right)±84}{2\left(-216\right)}$$

The opposite of -300 is 300.

$$k=\frac{300±84}{2\left(-216\right)}$$

Multiply 2 times -216.

$$k=\frac{300±84}{-432}$$

Now solve the equation k=\frac{300±84}{-432} when ± is plus. Add 300 to 84.

$$k=\frac{384}{-432}$$

Reduce the fraction \frac{384}{-432} to lowest terms by extracting and canceling out 48.

$$k=-\frac{8}{9}$$

Now solve the equation k=\frac{300±84}{-432} when ± is minus. Subtract 84 from 300.

$$k=\frac{216}{-432}$$

Reduce the fraction \frac{216}{-432} to lowest terms by extracting and canceling out 216.

$$k=-\frac{1}{2}$$

The equation is now solved.

$$k=-\frac{8}{9} k=-\frac{1}{2}$$

Steps for Completing the Square

Variable k cannot be equal to -\frac{2}{3} since division by zero is not defined. Multiply both sides of the equation by \left(3k+2\right)^{2}\left(9k^{2}+4\right), the least common multiple of 4+9k^{2}+12k,4+9k^{2}.

$$9k^{2}+4=\left(3k+2\right)^{2}\times 25$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3k+2\right)^{2}.

$$9k^{2}+4=\left(9k^{2}+12k+4\right)\times 25$$

Use the distributive property to multiply 9k^{2}+12k+4 by 25.

$$9k^{2}+4=225k^{2}+300k+100$$

Subtract 225k^{2} from both sides.

$$9k^{2}+4-225k^{2}=300k+100$$

Combine 9k^{2} and -225k^{2} to get -216k^{2}.

$$-216k^{2}+4=300k+100$$

Subtract 300k from both sides.

$$-216k^{2}+4-300k=100$$

Subtract 4 from both sides.

$$-216k^{2}-300k=100-4$$

Subtract 4 from 100 to get 96.

$$-216k^{2}-300k=96$$

Divide both sides by -216.

$$\frac{-216k^{2}-300k}{-216}=\frac{96}{-216}$$

Dividing by -216 undoes the multiplication by -216.

$$k^{2}+\left(-\frac{300}{-216}\right)k=\frac{96}{-216}$$

Reduce the fraction \frac{-300}{-216} to lowest terms by extracting and canceling out 12.

$$k^{2}+\frac{25}{18}k=\frac{96}{-216}$$

Reduce the fraction \frac{96}{-216} to lowest terms by extracting and canceling out 24.

$$k^{2}+\frac{25}{18}k=-\frac{4}{9}$$

Divide \frac{25}{18}, the coefficient of the x term, by 2 to get \frac{25}{36}. Then add the square of \frac{25}{36} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$k^{2}+\frac{25}{18}k+\left(\frac{25}{36}\right)^{2}=-\frac{4}{9}+\left(\frac{25}{36}\right)^{2}$$

Square \frac{25}{36} by squaring both the numerator and the denominator of the fraction.

$$k^{2}+\frac{25}{18}k+\frac{625}{1296}=-\frac{4}{9}+\frac{625}{1296}$$

Add -\frac{4}{9} to \frac{625}{1296} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$k^{2}+\frac{25}{18}k+\frac{625}{1296}=\frac{49}{1296}$$

Factor k^{2}+\frac{25}{18}k+\frac{625}{1296}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(k+\frac{25}{36}\right)^{2}=\frac{49}{1296}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(k+\frac{25}{36}\right)^{2}}=\sqrt{\frac{49}{1296}}$$

Simplify.

$$k+\frac{25}{36}=\frac{7}{36} k+\frac{25}{36}=-\frac{7}{36}$$

Subtract \frac{25}{36} from both sides of the equation.

$$k=-\frac{1}{2} k=-\frac{8}{9}$$