Solve 1+2k+1/4=9k^2/16

$$1+ \frac{ 2k+1 }{ 4 } = \frac{ 9 { k }^{ 2 } }{ 16 }$$

Solve for k

$$k = -\frac{10}{9} = -1\frac{1}{9} \approx -1.111111111$$

$$k=2$$

Steps Using Factoring By Grouping

Multiply both sides of the equation by 16, the least common multiple of 4,16.

$$16+4\left(2k+1\right)=9k^{2}$$

Use the distributive property to multiply 4 by 2k+1.

$$16+8k+4=9k^{2}$$

Add 16 and 4 to get 20.

$$20+8k=9k^{2}$$

Subtract 9k^{2} from both sides.

$$20+8k-9k^{2}=0$$

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

$$-9k^{2}+8k+20=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -9k^{2}+ak+bk+20. To find a and b, set up a system to be solved.

$$a+b=8 ab=-9\times 20=-180$$

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 -180.

$$-1,180 -2,90 -3,60 -4,45 -5,36 -6,30 -9,20 -10,18 -12,15$$

Calculate the sum for each pair.

$$-1+180=179 -2+90=88 -3+60=57 -4+45=41 -5+36=31 -6+30=24 -9+20=11 -10+18=8 -12+15=3$$

The solution is the pair that gives sum 8.

$$a=18 b=-10$$

Rewrite -9k^{2}+8k+20 as \left(-9k^{2}+18k\right)+\left(-10k+20\right).

$$\left(-9k^{2}+18k\right)+\left(-10k+20\right)$$

Factor out 9k in the first and 10 in the second group.

$$9k\left(-k+2\right)+10\left(-k+2\right)$$

Factor out common term -k+2 by using distributive property.

$$\left(-k+2\right)\left(9k+10\right)$$

To find equation solutions, solve -k+2=0 and 9k+10=0.

$$k=2 k=-\frac{10}{9}$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 16, the least common multiple of 4,16.

$$16+4\left(2k+1\right)=9k^{2}$$

Use the distributive property to multiply 4 by 2k+1.

$$16+8k+4=9k^{2}$$

Add 16 and 4 to get 20.

$$20+8k=9k^{2}$$

Subtract 9k^{2} from both sides.

$$20+8k-9k^{2}=0$$

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.

$$-9k^{2}+8k+20=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -9 for a, 8 for b, and 20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$k=\frac{-8±\sqrt{8^{2}-4\left(-9\right)\times 20}}{2\left(-9\right)}$$

Square 8.

$$k=\frac{-8±\sqrt{64-4\left(-9\right)\times 20}}{2\left(-9\right)}$$

Multiply -4 times -9.

$$k=\frac{-8±\sqrt{64+36\times 20}}{2\left(-9\right)}$$

Multiply 36 times 20.

$$k=\frac{-8±\sqrt{64+720}}{2\left(-9\right)}$$

Add 64 to 720.

$$k=\frac{-8±\sqrt{784}}{2\left(-9\right)}$$

Take the square root of 784.

$$k=\frac{-8±28}{2\left(-9\right)}$$

Multiply 2 times -9.

$$k=\frac{-8±28}{-18}$$

Now solve the equation k=\frac{-8±28}{-18} when ± is plus. Add -8 to 28.

$$k=\frac{20}{-18}$$

Reduce the fraction \frac{20}{-18} to lowest terms by extracting and canceling out 2.

$$k=-\frac{10}{9}$$

Now solve the equation k=\frac{-8±28}{-18} when ± is minus. Subtract 28 from -8.

$$k=-\frac{36}{-18}$$

Divide -36 by -18.

$$k=2$$

The equation is now solved.

$$k=-\frac{10}{9} k=2$$

Steps for Completing the Square

Multiply both sides of the equation by 16, the least common multiple of 4,16.

$$16+4\left(2k+1\right)=9k^{2}$$

Use the distributive property to multiply 4 by 2k+1.

$$16+8k+4=9k^{2}$$

Add 16 and 4 to get 20.

$$20+8k=9k^{2}$$

Subtract 9k^{2} from both sides.

$$20+8k-9k^{2}=0$$

Subtract 20 from both sides. Anything subtracted from zero gives its negation.

$$8k-9k^{2}=-20$$

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.

$$-9k^{2}+8k=-20$$

Divide both sides by -9.

$$\frac{-9k^{2}+8k}{-9}=-\frac{20}{-9}$$

Dividing by -9 undoes the multiplication by -9.

$$k^{2}+\frac{8}{-9}k=-\frac{20}{-9}$$

Divide 8 by -9.

$$k^{2}-\frac{8}{9}k=-\frac{20}{-9}$$

Divide -20 by -9.

$$k^{2}-\frac{8}{9}k=\frac{20}{9}$$

Divide -\frac{8}{9}, the coefficient of the x term, by 2 to get -\frac{4}{9}. Then add the square of -\frac{4}{9} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$k^{2}-\frac{8}{9}k+\left(-\frac{4}{9}\right)^{2}=\frac{20}{9}+\left(-\frac{4}{9}\right)^{2}$$

Square -\frac{4}{9} by squaring both the numerator and the denominator of the fraction.

$$k^{2}-\frac{8}{9}k+\frac{16}{81}=\frac{20}{9}+\frac{16}{81}$$

Add \frac{20}{9} to \frac{16}{81} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$k^{2}-\frac{8}{9}k+\frac{16}{81}=\frac{196}{81}$$

Factor k^{2}-\frac{8}{9}k+\frac{16}{81}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(k-\frac{4}{9}\right)^{2}=\frac{196}{81}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(k-\frac{4}{9}\right)^{2}}=\sqrt{\frac{196}{81}}$$

Simplify.

$$k-\frac{4}{9}=\frac{14}{9} k-\frac{4}{9}=-\frac{14}{9}$$

Add \frac{4}{9} to both sides of the equation.

$$k=2 k=-\frac{10}{9}$$