Solve q^2+q^3/1+q=9/4

$$\frac { q ^ { 2 } + q ^ { 3 } } { 1 + q } = \frac { 9 } { 4 }$$

Solve for q

$$q = -\frac{3}{2} = -1\frac{1}{2} = -1.5$$

$$q = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

Solution Steps

Variable q cannot be equal to -1 since division by zero is not defined. Multiply both sides of the equation by 4\left(q+1\right), the least common multiple of 1+q,4.

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

Use the distributive property to multiply 4 by q^{2}+q^{3}.

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

Use the distributive property to multiply 9 by q+1.

$$4q^{2}+4q^{3}=9q+9$$

Subtract 9q from both sides.

$$4q^{2}+4q^{3}-9q=9$$

Subtract 9 from both sides.

$$4q^{2}+4q^{3}-9q-9=0$$

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

$$4q^{3}+4q^{2}-9q-9=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -9 and q divides the leading coefficient 4. List all candidates \frac{p}{q}.

$$±\frac{9}{4},±\frac{9}{2},±9,±\frac{3}{4},±\frac{3}{2},±3,±\frac{1}{4},±\frac{1}{2},±1$$

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.

$$q=-1$$

By Factor theorem, q-k is a factor of the polynomial for each root k. Divide 4q^{3}+4q^{2}-9q-9 by q+1 to get 4q^{2}-9. Solve the equation where the result equals to 0.

$$4q^{2}-9=0$$

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 4 for a, 0 for b, and -9 for c in the quadratic formula.

$$q=\frac{0±\sqrt{0^{2}-4\times 4\left(-9\right)}}{2\times 4}$$

Do the calculations.

$$q=\frac{0±12}{8}$$

Solve the equation 4q^{2}-9=0 when ± is plus and when ± is minus.

$$q=-\frac{3}{2} q=\frac{3}{2}$$

Remove the values that the variable cannot be equal to.

$$q=-\frac{3}{2}\text{ or }q=\frac{3}{2}$$

List all found solutions.

$$q=-1 q=-\frac{3}{2} q=\frac{3}{2}$$

Variable q cannot be equal to -1.

$$q=\frac{3}{2} q=-\frac{3}{2}$$