Solve 9k-3=4k(3k-1)/1+4k^2

$$9 k - 3 = \frac { 4 k ( 3 k - 1 ) } { 1 + 4 k ^ { 2 } }$$

Solve for k

$$k=\frac{1}{3}\approx 0.333333333$$

Solution Steps

Multiply both sides of the equation by 4k^{2}+1.

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

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

$$36k^{3}+9k+\left(4k^{2}+1\right)\left(-3\right)=4k\left(3k-1\right)$$

Use the distributive property to multiply 4k^{2}+1 by -3.

$$36k^{3}+9k-12k^{2}-3=4k\left(3k-1\right)$$

Use the distributive property to multiply 4k by 3k-1.

$$36k^{3}+9k-12k^{2}-3=12k^{2}-4k$$

Subtract 12k^{2} from both sides.

$$36k^{3}+9k-12k^{2}-3-12k^{2}=-4k$$

Combine -12k^{2} and -12k^{2} to get -24k^{2}.

$$36k^{3}+9k-24k^{2}-3=-4k$$

Add 4k to both sides.

$$36k^{3}+9k-24k^{2}-3+4k=0$$

Combine 9k and 4k to get 13k.

$$36k^{3}+13k-24k^{2}-3=0$$

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

$$36k^{3}-24k^{2}+13k-3=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -3 and q divides the leading coefficient 36. List all candidates \frac{p}{q}.

$$±\frac{1}{12},±\frac{1}{6},±\frac{1}{4},±\frac{1}{3},±\frac{1}{2},±\frac{3}{4},±1,±\frac{3}{2},±3,±\frac{1}{36},±\frac{1}{18},±\frac{1}{9}$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$k=\frac{1}{3}$$

By Factor theorem, k-k is a factor of the polynomial for each root k. Divide 36k^{3}-24k^{2}+13k-3 by 3\left(k-\frac{1}{3}\right)=3k-1 to get 12k^{2}-4k+3. Solve the equation where the result equals to 0.

$$12k^{2}-4k+3=0$$

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 12 for a, -4 for b, and 3 for c in the quadratic formula.

$$k=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 12\times 3}}{2\times 12}$$

Do the calculations.

$$k=\frac{4±\sqrt{-128}}{24}$$

Since the square root of a negative number is not defined in the real field, there are no solutions.

$$k\in \emptyset$$

List all found solutions.

$$k=\frac{1}{3}$$