Problem 48 - College Algebra

Question

Solve each equation in by making an appropriate substitution. $$ 2 x^{2 / 3}+7 x^{1 / 3}-15=0 $$

Step-by-Step Solution

Answer

The solutions to given equation are \(x = \frac{27}{8}\) and \(x = -125\)

Step 1: Variable Substitution

First, identify the smallest power of \(x\), which in our case is \(\frac{1}{3}\). Now, let's set a new variable such that \(u = x^{1 / 3}\). The equation can then be rewritten in terms of \(u\) as \(2u^2 + 7u - 15 = 0 \)

Step 2: Solving the Equation

This equation is in quadratic form. It can be solved by factorizing or by using the quadratic formula. Here, \(2u^2 + 7u - 15\) can be factored as \((2u - 3)(u + 5) = 0\). By the 'zero-product property', if a product of two factors equals zero, then at least one of the factors must be equal to zero. Therefore, we get two possible solutions for \(u\): \(u = \frac{3}{2}\) or \(u = -5\)

Step 3: Replacing the substituted variable

Now we replace \(u\) with \(x^{1 / 3}\) to get the solutions for original variable \(x\). Hence \(x^{1 / 3} = \frac{3}{2}\) or \(x^{1 / 3} = -5\). Cubing both sides for each of these solutions provides us: \(x = (\frac{3}{2})^3 = \frac{27}{8}\) or \(x = -5^3 = -125\)