# i need help solving this problem 64d^3=100d?

Feb 16, 2017

$d = 0 \text{ or } d = \pm \frac{5}{4}$

#### Explanation:

Rearrange the equation and equate to zero.

$\Rightarrow 64 {d}^{3} - 100 d = 0$

Take out a $\textcolor{b l u e}{\text{common factor}}$ of 4d

$\Rightarrow 4 d \left(16 {d}^{2} - 25\right) = 0 \rightarrow \left(1\right)$

16d^2-25" is a $\textcolor{b l u e}{\text{difference of squares}}$ and factorises, in general, as follows.

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{{a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

$\text{here "a=4d" and } b = 5$

$\Rightarrow 16 {d}^{2} - 25 = \left(4 d - 5\right) \left(4 d + 5\right)$

Returning to (1) and completing.

$\Rightarrow 4 d \left(4 d - 5\right) \left(4 d + 5\right) = 0$

Solving each of the 3 factors.

$4 d = 0 \Rightarrow d = 0$

$4 d - 5 = 0 \Rightarrow d = \frac{5}{4}$

$4 d + 5 = 0 \Rightarrow d = - \frac{5}{4}$

$\Rightarrow d = 0 \text{ or " d=+-5/4" are the solutions}$