Question

question
solve algebraically for all values of x:
$x^5 - 7x^3 = 0$
answer attempt 1 out of 2
additional solution no solution
$x = $
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Explanation:

Step1: Factor the equation

Factor out the greatest common factor \(x^3\) from the left - hand side of the equation \(x^{5}-7x^{3}=0\). We get \(x^{3}(x^{2}-7)=0\).

Step2: Apply the zero - product property

The zero - product property states that if \(ab = 0\), then either \(a = 0\) or \(b = 0\). So, we set each factor equal to zero:

• Case 1: \(x^{3}=0\)

Taking the cube root of both sides, we have \(x=\sqrt[3]{0}=0\).

• Case 2: \(x^{2}-7 = 0\)

Add 7 to both sides of the equation: \(x^{2}=7\).
Then take the square root of both sides. Remember that if \(x^{2}=a\) (\(a\geq0\)), then \(x=\pm\sqrt{a}\). So, \(x=\pm\sqrt{7}\).

Answer:

\(x = 0\), \(x=\sqrt{7}\), \(x =-\sqrt{7}\)