Question
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2) \( x^{2 / 3}-2 x^{1 / 3}-3=0 \)

Ask by Osborne Paul. in South Africa
Mar 14,2025

Upstudy AI Solution

Tutor-Verified Answer

Answer

The solutions are \(x=-1\) and \(x=27\).

Solution

Solve the equation \( x^{2 / 3}-2x^{1 / 3}-3=0 \). Solve the equation by following steps: - step0: Solve for \(x\): \(x^{\frac{2}{3}}-2x^{\frac{1}{3}}-3=0\) - step1: Solve using substitution: \(t^{2}-2t-3=0\) - step2: Factor the expression: \(\left(t-3\right)\left(t+1\right)=0\) - step3: Separate into possible cases: \(\begin{align}&t-3=0\\&t+1=0\end{align}\) - step4: Solve the equation: \(\begin{align}&t=3\\&t=-1\end{align}\) - step5: Substitute back: \(\begin{align}&x^{\frac{1}{3}}=3\\&x^{\frac{1}{3}}=-1\end{align}\) - step6: Solve the equation for \(x:\) \(\begin{align}&x=27\\&x=-1\end{align}\) - step7: Rewrite: \(x_{1}=-1,x_{2}=27\) The solutions to the equation \(x^{2 / 3}-2x^{1 / 3}-3=0\) are \(x=-1\) and \(x=27\).

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Beyond the Answer

To solve the equation \( x^{2/3} - 2x^{1/3} - 3 = 0 \), let's make a substitution. Let \( y = x^{1/3} \). Then, the equation transforms into \( y^2 - 2y - 3 = 0 \). This quadratic can be factored or solved using the quadratic formula. Factoring gives us \( (y - 3)(y + 1) = 0 \), leading to solutions \( y = 3 \) and \( y = -1 \). Reverting back to \( x \), we have \( x^{1/3} = 3 \) resulting in \( x = 27 \), and \( x^{1/3} = -1 \) giving \( x = -1 \). Thus, the solutions are \( x = 27 \) and \( x = -1 \). Now, let's check the solutions to ensure they're valid in the original equation. Plugging \( x = 27 \) back in yields \( 27^{2/3} - 2(27^{1/3}) - 3 = 9 - 6 - 3 = 0 \), confirming that it's correct. Next, for \( x = -1 \): \( (-1)^{2/3} - 2(-1)^{1/3} - 3 = 1 + 2 - 3 = 0 \), which also checks out. Hence, both solutions are valid!

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