Question

Solve 2x3−24x2+54x=0

Place roots in order from least to greatest.

(3 points)
The roots are x=

, x=
, x=

Answer 1

To solve the equation \( 2x^3 - 24x^2 + 54x = 0 \), we can start by factoring out the common term:

\[
2x(x^2 - 12x + 27) = 0
\]

This gives us one root:

\[
2x = 0 \implies x = 0
\]

Now we need to solve the quadratic \( x^2 - 12x + 27 = 0 \). We can use the quadratic formula for this:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Here, \( a = 1 \), \( b = -12 \), and \( c = 27 \).

Calculating the discriminant:

\[
b^2 - 4ac = (-12)^2 - 4 \cdot 1 \cdot 27 = 144 - 108 = 36
\]

Now, using the quadratic formula:

\[
x = \frac{12 \pm \sqrt{36}}{2 \cdot 1} = \frac{12 \pm 6}{2}
\]

Calculating the two possible roots:

1. \( x = \frac{12 + 6}{2} = \frac{18}{2} = 9 \)
2. \( x = \frac{12 - 6}{2} = \frac{6}{2} = 3 \)

Thus, the roots of the equation \( 2x^3 - 24x^2 + 54x = 0 \) are:

\[
x = 0, x = 3, x = 9
\]

Placing the roots in order from least to greatest, we have:

\[
x = 0, x = 3, x = 9
\]

Therefore, the final answer is:

The roots are \( x = 0 \), \( x = 3 \), \( x = 9 \).