Answer :
To solve the polynomial equation [tex]\(25x^3 - 64x = 0\)[/tex], follow these steps:
1. Factor the Equation:
Start by factoring out the common term, which is [tex]\(x\)[/tex]:
[tex]\[
x(25x^2 - 64) = 0
\][/tex]
This gives us two potential solutions: when [tex]\(x = 0\)[/tex] or when the expression in the parentheses equals zero.
2. Solve for [tex]\(x = 0\)[/tex]:
One of the solutions is straightforward:
[tex]\[
x = 0
\][/tex]
3. Solve the Quadratic Equation:
Now, solve the quadratic equation [tex]\(25x^2 - 64 = 0\)[/tex].
a. Rearrange the Equation:
[tex]\[
25x^2 = 64
\][/tex]
b. Isolate [tex]\(x^2\)[/tex]:
[tex]\[
x^2 = \frac{64}{25}
\][/tex]
c. Take the Square Root:
Find the square root of both sides:
[tex]\[
x = \pm \sqrt{\frac{64}{25}}
\][/tex]
This simplifies to:
[tex]\[
x = \pm \frac{8}{5}
\][/tex]
Calculating gives:
[tex]\[
x = 1.6 \quad \text{or} \quad x = -1.6
\][/tex]
4. Solutions:
The complete set of solutions for the equation [tex]\(25x^3 - 64x = 0\)[/tex] are:
[tex]\[
x = 0, \quad x = 1.6, \quad x = -1.6
\][/tex]
These are the values of [tex]\(x\)[/tex] that satisfy the original equation.
1. Factor the Equation:
Start by factoring out the common term, which is [tex]\(x\)[/tex]:
[tex]\[
x(25x^2 - 64) = 0
\][/tex]
This gives us two potential solutions: when [tex]\(x = 0\)[/tex] or when the expression in the parentheses equals zero.
2. Solve for [tex]\(x = 0\)[/tex]:
One of the solutions is straightforward:
[tex]\[
x = 0
\][/tex]
3. Solve the Quadratic Equation:
Now, solve the quadratic equation [tex]\(25x^2 - 64 = 0\)[/tex].
a. Rearrange the Equation:
[tex]\[
25x^2 = 64
\][/tex]
b. Isolate [tex]\(x^2\)[/tex]:
[tex]\[
x^2 = \frac{64}{25}
\][/tex]
c. Take the Square Root:
Find the square root of both sides:
[tex]\[
x = \pm \sqrt{\frac{64}{25}}
\][/tex]
This simplifies to:
[tex]\[
x = \pm \frac{8}{5}
\][/tex]
Calculating gives:
[tex]\[
x = 1.6 \quad \text{or} \quad x = -1.6
\][/tex]
4. Solutions:
The complete set of solutions for the equation [tex]\(25x^3 - 64x = 0\)[/tex] are:
[tex]\[
x = 0, \quad x = 1.6, \quad x = -1.6
\][/tex]
These are the values of [tex]\(x\)[/tex] that satisfy the original equation.