Solve frac{6(k^{2}+1)^{2}-(3k^2-1)^2}{(3k^2+1)^2}=5/4

$$\frac { 6 ( k ^ { 2 } + 1 ) ^ { 2 } - ( 3 k ^ { 2 } - 1 ) ^ { 2 } } { ( 3 k ^ { 2 } + 1 ) ^ { 2 } } = \frac { 5 } { 4 }$$

Solve for k

$$k=-1$$

$$k=1$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 4\left(3k^{2}+1\right)^{2}, the least common multiple of \left(3k^{2}+1\right)^{2},4.

$$4\left(6\left(k^{2}+1\right)^{2}-\left(3k^{2}-1\right)^{2}\right)=5\left(3k^{2}+1\right)^{2}$$

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

$$4\left(6\left(\left(k^{2}\right)^{2}+2k^{2}+1\right)-\left(3k^{2}-1\right)^{2}\right)=5\left(3k^{2}+1\right)^{2}$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$4\left(6\left(k^{4}+2k^{2}+1\right)-\left(3k^{2}-1\right)^{2}\right)=5\left(3k^{2}+1\right)^{2}$$

Use the distributive property to multiply 6 by k^{4}+2k^{2}+1.

$$4\left(6k^{4}+12k^{2}+6-\left(3k^{2}-1\right)^{2}\right)=5\left(3k^{2}+1\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3k^{2}-1\right)^{2}.

$$4\left(6k^{4}+12k^{2}+6-\left(9\left(k^{2}\right)^{2}-6k^{2}+1\right)\right)=5\left(3k^{2}+1\right)^{2}$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$4\left(6k^{4}+12k^{2}+6-\left(9k^{4}-6k^{2}+1\right)\right)=5\left(3k^{2}+1\right)^{2}$$

To find the opposite of 9k^{4}-6k^{2}+1, find the opposite of each term.

$$4\left(6k^{4}+12k^{2}+6-9k^{4}+6k^{2}-1\right)=5\left(3k^{2}+1\right)^{2}$$

Combine 6k^{4} and -9k^{4} to get -3k^{4}.

$$4\left(-3k^{4}+12k^{2}+6+6k^{2}-1\right)=5\left(3k^{2}+1\right)^{2}$$

Combine 12k^{2} and 6k^{2} to get 18k^{2}.

$$4\left(-3k^{4}+18k^{2}+6-1\right)=5\left(3k^{2}+1\right)^{2}$$

Subtract 1 from 6 to get 5.

$$4\left(-3k^{4}+18k^{2}+5\right)=5\left(3k^{2}+1\right)^{2}$$

Use the distributive property to multiply 4 by -3k^{4}+18k^{2}+5.

$$-12k^{4}+72k^{2}+20=5\left(3k^{2}+1\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3k^{2}+1\right)^{2}.

$$-12k^{4}+72k^{2}+20=5\left(9\left(k^{2}\right)^{2}+6k^{2}+1\right)$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$-12k^{4}+72k^{2}+20=5\left(9k^{4}+6k^{2}+1\right)$$

Use the distributive property to multiply 5 by 9k^{4}+6k^{2}+1.

$$-12k^{4}+72k^{2}+20=45k^{4}+30k^{2}+5$$

Subtract 45k^{4} from both sides.

$$-12k^{4}+72k^{2}+20-45k^{4}=30k^{2}+5$$

Combine -12k^{4} and -45k^{4} to get -57k^{4}.

$$-57k^{4}+72k^{2}+20=30k^{2}+5$$

Subtract 30k^{2} from both sides.

$$-57k^{4}+72k^{2}+20-30k^{2}=5$$

Combine 72k^{2} and -30k^{2} to get 42k^{2}.

$$-57k^{4}+42k^{2}+20=5$$

Subtract 5 from both sides.

$$-57k^{4}+42k^{2}+20-5=0$$

Subtract 5 from 20 to get 15.

$$-57k^{4}+42k^{2}+15=0$$

Substitute t for k^{2}.

$$-57t^{2}+42t+15=0$$

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 -57 for a, 42 for b, and 15 for c in the quadratic formula.

$$t=\frac{-42±\sqrt{42^{2}-4\left(-57\right)\times 15}}{-57\times 2}$$

Do the calculations.

$$t=\frac{-42±72}{-114}$$

Solve the equation t=\frac{-42±72}{-114} when ± is plus and when ± is minus.

$$t=-\frac{5}{19} t=1$$

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

$$k=1 k=-1$$

Solve for k (complex solution)

$$k=\frac{\sqrt{95}i}{19}\approx 0.512989176i$$

$$k=-\frac{\sqrt{95}i}{19}\approx -0-0.512989176i$$

$$k=-1$$

$$k=1$$