Q. 435 - Intermediate Algebra

Question

In the following exercises, solve.

$16p^3=24p^2-9p$

Step-by-Step Solution

Answer

Answer is $p=0,p=\frac{3}{4}$

Step 1. Given

The equation is $16p^3=24p^2-9p$

Step 2. Simplify the equation

$$\begin{gathered} 16p^3=24p^2-9p \\ 16p^3-24p^2+9p=0 \\ p(16p^2-24p+9)=0 \\ p{(4p-3)}^2=0 \end{gathered}$$

So,

$p=0$

and

$$\begin{gathered} 4p-3=0 \\ p=\frac{3}{4} \end{gathered}$$

Step 3. Check both the values

Put $p=0$ in the original equation,

$$\begin{gathered} 16p^3=24p^2-9p \\ 16{\left(0\right)}^3=24{\left(0\right)}^2-9\left(0\right) \\ 0=0 \end{gathered}$$

True

Now put $p=\frac{3}{4}$ in the original equation,

$$\begin{gathered} 16p^3=24p^2-9p \\ 16{\left(\frac{3}{4}\right)}^3=24{\left(\frac{3}{4}\right)}^2-9\left(\frac{3}{4}\right) \\ 16\times \frac{27}{64}=24\times \frac{9}{16}-\frac{27}{4} \\ \frac{27}{4}=\frac{54}{4}-\frac{27}{4} \\ \frac{27}{54}=\frac{27}{54} \end{gathered}$$

True

Step 4. Required answer

Answer is $p=0,p=\frac{3}{4}$