Solve 9k^2(k^2+3/4)=4-4k^2

$$9 k ^ { 2 } ( k ^ { 2 } + \frac { 3 } { 4 } ) = 4 - 4 k ^ { 2 }$$

Solve for k

$$k=-\frac{\sqrt{2\left(\sqrt{4153}-43\right)}}{12}\approx -0.54573821$$

$$k=\frac{\sqrt{2\left(\sqrt{4153}-43\right)}}{12}\approx 0.54573821$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 9k^{2} by k^{2}+\frac{3}{4}.

$$9k^{4}+\frac{27}{4}k^{2}=4-4k^{2}$$

Subtract 4 from both sides.

$$9k^{4}+\frac{27}{4}k^{2}-4=-4k^{2}$$

Add 4k^{2} to both sides.

$$9k^{4}+\frac{27}{4}k^{2}-4+4k^{2}=0$$

Combine \frac{27}{4}k^{2} and 4k^{2} to get \frac{43}{4}k^{2}.

$$9k^{4}+\frac{43}{4}k^{2}-4=0$$

Substitute t for k^{2}.

$$9t^{2}+\frac{43}{4}t-4=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 9 for a, \frac{43}{4} for b, and -4 for c in the quadratic formula.

$$t=\frac{-\frac{43}{4}±\sqrt{\left(\frac{43}{4}\right)^{2}-4\times 9\left(-4\right)}}{2\times 9}$$

Do the calculations.

$$t=\frac{-\frac{43}{4}±\frac{1}{4}\sqrt{4153}}{18}$$

Solve the equation t=\frac{-\frac{43}{4}±\frac{1}{4}\sqrt{4153}}{18} when ± is plus and when ± is minus.

$$t=\frac{\sqrt{4153}-43}{72} t=\frac{-\sqrt{4153}-43}{72}$$

Since k=t^{2}, the solutions are obtained by evaluating k=±\sqrt{t} for positive t.

$$k=\frac{\sqrt{\frac{\sqrt{4153}-43}{2}}}{6} k=-\frac{\sqrt{\frac{\sqrt{4153}-43}{2}}}{6}$$