Question

solving polynomial equations by factoring
1/3h^2+13/12h=-11/8h-7/8

Answer 1

\[\frac{1}{3 h^2}+\frac{13}{12 h}=-\frac{11}{8 h}-\frac{7}{8}\]????????????

Answer 2

the h^2 suppose to be between 1/3 and the h is suppose to be 13/12 and 11/8

Answer 3

\[\frac{1}{3} h^2+\frac{13}{12} h=-\frac{11}{8} h-\frac{7}{8}\]

Answer 4

yes that specific way

Answer 5

\[\begin{array}{l} \frac{h^2}{3}+\frac{13 h}{12}=-\frac{11 h}{8}-\frac{7}{8} \\ \text{Subtract }\left(-\frac{11 h}{8}-\frac{7}{8}\right)\text{ from both sides:} \\ \frac{h^2}{3}+\frac{59 h}{24}+\frac{7}{8}=0 \\ \text{Solve the quadratic equation by completing the \square:}\text{$\backslash $n$\backslash $n}\text{Divide both sides by }\frac{1}{3}: \\ h^2+\frac{59 h}{8}+\frac{21}{8}=0 \\ \text{Subtract }\frac{21}{8}\text{ from both sides:} \\ h^2+\frac{59 h}{8}=-\frac{21}{8} \\ \text{Add }\frac{3481}{256}\text{ \to both sides:} \\ h^2+\frac{59 h}{8}+\frac{3481}{256}=\frac{2809}{256} \\ \text{Factor the \left hand side:} \\ \left(h+\frac{59}{16}\right)^2=\frac{2809}{256} \\ \text{Take the \square \root of both sides:} \\ \left|h+\frac{59}{16}\right|=\frac{53}{16} \\ \text{Eliminate the absolute value:} \\ h+\frac{59}{16}=-\frac{53}{16}\text{ or }h+\frac{59}{16}=\frac{53}{16} \\ \text{Subtract }\frac{59}{16}\text{ from both sides:} \\ h=-7\text{ or }h+\frac{59}{16}=\frac{53}{16} \\ \text{Subtract }\frac{59}{16}\text{ from both sides:} \\ h=-7\text{ or }h=-\frac{3}{8} \\\end{array}\]

Answer 6

is there a easier way i can understand and learn to solve this problem?