Solve 28r^3+7r=40r^2+10

$$28 r ^ { 3 } + 7 r = 40 r ^ { 2 } + 10$$

Solve for r

$$r = \frac{10}{7} = 1\frac{3}{7} \approx 1.428571429$$

Solution Steps

Subtract 40r^{2} from both sides.

$$28r^{3}+7r-40r^{2}=10$$

Subtract 10 from both sides.

$$28r^{3}+7r-40r^{2}-10=0$$

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

$$28r^{3}-40r^{2}+7r-10=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -10 and q divides the leading coefficient 28. List all candidates \frac{p}{q}.

$$±\frac{5}{14},±\frac{5}{7},±\frac{10}{7},±\frac{5}{2},±5,±10,±\frac{5}{28},±\frac{5}{4},±\frac{1}{14},±\frac{1}{7},±\frac{2}{7},±\frac{1}{2},±1,±2,±\frac{1}{28},±\frac{1}{4}$$

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.

$$r=\frac{10}{7}$$

By Factor theorem, r-k is a factor of the polynomial for each root k. Divide 28r^{3}-40r^{2}+7r-10 by 7\left(r-\frac{10}{7}\right)=7r-10 to get 4r^{2}+1. Solve the equation where the result equals to 0.

$$4r^{2}+1=0$$

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 1 for c in the quadratic formula.

$$r=\frac{0±\sqrt{0^{2}-4\times 4\times 1}}{2\times 4}$$

Do the calculations.

$$r=\frac{0±\sqrt{-16}}{8}$$

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

$$r\in \emptyset$$

List all found solutions.

$$r=\frac{10}{7}$$

Solve for r (complex solution)

$$r=-\frac{1}{2}i=-0.5i$$

$$r = \frac{10}{7} = 1\frac{3}{7} \approx 1.428571429$$

$$r=\frac{1}{2}i=0.5i$$