Solve (9k-27)^2=(2k+4k+27)*81/2

$$( 9 k - 27 ) ^ { 2 } = ( 2 k + 4 k + 27 ) \cdot \frac { 81 } { 2 }$$

Solve for k

$$k = \frac{3 \sqrt{11} + 9}{2} \approx 9.474937186$$

$$k=\frac{9-3\sqrt{11}}{2}\approx -0.474937186$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(9k-27\right)^{2}.

$$81k^{2}-486k+729=\left(2k+4k+27\right)\times \frac{81}{2}$$

Combine 2k and 4k to get 6k.

$$81k^{2}-486k+729=\left(6k+27\right)\times \frac{81}{2}$$

Use the distributive property to multiply 6k+27 by \frac{81}{2}.

$$81k^{2}-486k+729=243k+\frac{2187}{2}$$

Subtract 243k from both sides.

$$81k^{2}-486k+729-243k=\frac{2187}{2}$$

Combine -486k and -243k to get -729k.

$$81k^{2}-729k+729=\frac{2187}{2}$$

Subtract \frac{2187}{2} from both sides.

$$81k^{2}-729k+729-\frac{2187}{2}=0$$

Subtract \frac{2187}{2} from 729 to get -\frac{729}{2}.

$$81k^{2}-729k-\frac{729}{2}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 81 for a, -729 for b, and -\frac{729}{2} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$k=\frac{-\left(-729\right)±\sqrt{\left(-729\right)^{2}-4\times 81\left(-\frac{729}{2}\right)}}{2\times 81}$$

Square -729.

$$k=\frac{-\left(-729\right)±\sqrt{531441-4\times 81\left(-\frac{729}{2}\right)}}{2\times 81}$$

Multiply -4 times 81.

$$k=\frac{-\left(-729\right)±\sqrt{531441-324\left(-\frac{729}{2}\right)}}{2\times 81}$$

Multiply -324 times -\frac{729}{2}.

$$k=\frac{-\left(-729\right)±\sqrt{531441+118098}}{2\times 81}$$

Add 531441 to 118098.

$$k=\frac{-\left(-729\right)±\sqrt{649539}}{2\times 81}$$

Take the square root of 649539.

$$k=\frac{-\left(-729\right)±243\sqrt{11}}{2\times 81}$$

The opposite of -729 is 729.

$$k=\frac{729±243\sqrt{11}}{2\times 81}$$

Multiply 2 times 81.

$$k=\frac{729±243\sqrt{11}}{162}$$

Now solve the equation k=\frac{729±243\sqrt{11}}{162} when ± is plus. Add 729 to 243\sqrt{11}.

$$k=\frac{243\sqrt{11}+729}{162}$$

Divide 729+243\sqrt{11} by 162.

$$k=\frac{3\sqrt{11}+9}{2}$$

Now solve the equation k=\frac{729±243\sqrt{11}}{162} when ± is minus. Subtract 243\sqrt{11} from 729.

$$k=\frac{729-243\sqrt{11}}{162}$$

Divide 729-243\sqrt{11} by 162.

$$k=\frac{9-3\sqrt{11}}{2}$$

The equation is now solved.

$$k=\frac{3\sqrt{11}+9}{2} k=\frac{9-3\sqrt{11}}{2}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(9k-27\right)^{2}.

$$81k^{2}-486k+729=\left(2k+4k+27\right)\times \frac{81}{2}$$

Combine 2k and 4k to get 6k.

$$81k^{2}-486k+729=\left(6k+27\right)\times \frac{81}{2}$$

Use the distributive property to multiply 6k+27 by \frac{81}{2}.

$$81k^{2}-486k+729=243k+\frac{2187}{2}$$

Subtract 243k from both sides.

$$81k^{2}-486k+729-243k=\frac{2187}{2}$$

Combine -486k and -243k to get -729k.

$$81k^{2}-729k+729=\frac{2187}{2}$$

Subtract 729 from both sides.

$$81k^{2}-729k=\frac{2187}{2}-729$$

Subtract 729 from \frac{2187}{2} to get \frac{729}{2}.

$$81k^{2}-729k=\frac{729}{2}$$

Divide both sides by 81.

$$\frac{81k^{2}-729k}{81}=\frac{\frac{729}{2}}{81}$$

Dividing by 81 undoes the multiplication by 81.

$$k^{2}+\left(-\frac{729}{81}\right)k=\frac{\frac{729}{2}}{81}$$

Divide -729 by 81.

$$k^{2}-9k=\frac{\frac{729}{2}}{81}$$

Divide \frac{729}{2} by 81.

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

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

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

Square -\frac{9}{2} by squaring both the numerator and the denominator of the fraction.

$$k^{2}-9k+\frac{81}{4}=\frac{9}{2}+\frac{81}{4}$$

Add \frac{9}{2} to \frac{81}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$k^{2}-9k+\frac{81}{4}=\frac{99}{4}$$

Factor k^{2}-9k+\frac{81}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(k-\frac{9}{2}\right)^{2}=\frac{99}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(k-\frac{9}{2}\right)^{2}}=\sqrt{\frac{99}{4}}$$

Simplify.

$$k-\frac{9}{2}=\frac{3\sqrt{11}}{2} k-\frac{9}{2}=-\frac{3\sqrt{11}}{2}$$

Add \frac{9}{2} to both sides of the equation.

$$k=\frac{3\sqrt{11}+9}{2} k=\frac{9-3\sqrt{11}}{2}$$