Answered

Solve the cube root equation $\sqrt[3]{2 x^2}=\sqrt[3]{3-5 x}$.


A$x=-\frac{5}{2}$ and $x=1$
B$x=1$ and $x=\frac{3}{2}$
C$x=\frac{5}{2}$ and $x=1$
D$x=-3$ and $x=\frac{1}{2}$

Verified Answer :

  • Eliminate the cube roots by equating the internal polynomials: \(2x^2 = 3 - 5x\).
  • Rearrange into a standard quadratic form: \(2x^2 + 5x - 3 = 0\).
  • Apply the quadratic formula to find the roots: \(x = \frac{-5 \pm 7}{4}\).
  • Identify the solution set: \(\boxed{x = -3, x = \frac{1}{2}}\).

Explanation

  1. Equating and Simplifying the Expressions

To solve the equation \(\sqrt[3]{2x^2} = \sqrt[3]{3-5x}\), we utilize the fact that the cube root function is injective, meaning that if \(\sqrt[3]{u} = \sqrt[3]{v}\), then it must be that \(u = v\). Applying this property to the given equation allows us to equate the expressions inside the cube roots:
\[2x^2 = 3 - 5x\]
This is a standard quadratic equation. To solve it, we move all terms to one side of the equation to set it to zero:
\[2x^2 + 5x - 3 = 0\]

  1. Solving the Quadratic Equation

We now solve the quadratic equation \(2x^2 + 5x - 3 = 0\) using the quadratic formula, where \(a = 2\), \(b = 5\), and \(c = -3\):
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
Substituting the values of \(a, b,\) and \(c\):
\[x = \frac{-5 \pm \sqrt{5^2 - 4 \times 2 \times (-3)}}{2 \times 2}\]
\[x = \frac{-5 \pm \sqrt{25 + 24}}{4}\]
\[x = \frac{-5 \pm \sqrt{49}}{4}\]
\[x = \frac{-5 \pm 7}{4}\]
This yields two possible values for \(x\):
\[x_1 = \frac{-5 + 7}{4} = \frac{2}{4} = \frac{1}{2}\]
\[x_2 = \frac{-5 - 7}{4} = \frac{-12}{4} = -3\]
Thus, the solutions are \(x = \frac{1}{2}\) and \(x = -3\). Comparing this result with the provided options, it matches the last choice.

Examples

This type of problem is foundational in algebraic modeling, such as finding the equilibrium points in economic models where two different growth or cost functions (like those involving power terms) intersect. For example, if you are analyzing the supply and demand curves where quantities follow power relationships, solving for the intersection point helps determine the exact price point where the system balances. Finding such intersection points is a critical skill for understanding how different mathematical models interact and reach stability.

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