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Question
9x^{3}=4x^{5}
Solve the equation
x_{1}=-\frac{3}{2},x_{2}=0,x_{3}=\frac{3}{2}
Alternative Form
x_{1}=-1.5,x_{2}=0,x_{3}=1.5
Evaluate
9x^{3}=4x^{5}
Add or subtract both sides
9x^{3}-4x^{5}=0
Factor the expression
x^{3}\left(9-4x^{2}\right)=0
\text{Separate the equation into }2\text{ possible cases}
\begin{align}&x^{3}=0\\&9-4x^{2}=0\end{align}
The only way a power can be 0 is when the base equals 0
\begin{align}&x=0\\&9-4x^{2}=0\end{align}
Solve the equation
More Steps
Evaluate
9-4x^{2}=0
Move the constant to the right-hand side and change its sign
-4x^{2}=0-9
Removing 0 doesn't change the value,so remove it from the expression
-4x^{2}=-9
Change the signs on both sides of the equation
4x^{2}=9
Divide both sides
\frac{4x^{2}}{4}=\frac{9}{4}
Divide the numbers
x^{2}=\frac{9}{4}
Take the root of both sides of the equation and remember to use both positive and negative roots
x=\pm \sqrt{\frac{9}{4}}
Simplify the expression
More Steps
Evaluate
\sqrt{\frac{9}{4}}
To take a root of a fraction,take the root of the numerator and denominator separately
\frac{\sqrt{9}}{\sqrt{4}}
Simplify the radical expression
\frac{3}{\sqrt{4}}
Simplify the radical expression
\frac{3}{2}
x=\pm \frac{3}{2}
\text{Separate the equation into }2\text{ possible cases}
\begin{align}&x=\frac{3}{2}\\&x=-\frac{3}{2}\end{align}
\begin{align}&x=0\\&x=\frac{3}{2}\\&x=-\frac{3}{2}\end{align}
Solution
x_{1}=-\frac{3}{2},x_{2}=0,x_{3}=\frac{3}{2}
Alternative Form
x_{1}=-1.5,x_{2}=0,x_{3}=1.5
Show Solutions
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