Question

solve the equation.

\\x - 14\sqrt{x} - 32 = 0\\

select the correct choice below and, if necessary, fill in the

a. the solution set is \\{\quad\\}.
(simplify your answer. type an exact answer, using

b. the solution set is the empty set.

Explanation:

🆕 New Concept Discovered: Equations Quadratic in Form
Solving equations by substituting a new variable to reveal a quadratic structure.

Step 1: Identify the quadratic structure

We are given the equation:

$$ x - 14\sqrt{x} - 32 = 0 $$

Notice that the first term \( x \) is the square of the middle variable term \(\sqrt{x}\), since \((\sqrt{x})^2 = x\) (for \(x \ge 0\)). This means we can rewrite the equation in a quadratic form by substituting a new variable.

Let:

$$ u = \sqrt{x} $$

Substituting \( u \) and \( u^2 = x \) into the original equation gives:

$$ u^2 - 14u - 32 = 0 $$

Step 2: Solve the quadratic equation

Now, we solve for \( u \) by factoring the quadratic equation. We need two numbers that multiply to \(-32\) and add up to \(-14\). These numbers are \(-16\) and \(2\).

$$ (u - 16)(u + 2) = 0 $$

Setting each factor to zero gives:

$$ u = 16 \quad \text{or} \quad u = -2 $$

Step 3: Substitute back and solve for x

Now we replace \( u \) back with \(\sqrt{x}\) to find the values of \( x \).

1. For \( u = 16 \):

$$ \sqrt{x} = 16 $$

Squaring both sides:

$$ x = 16^2 = 256 $$

1. For \( u = -2 \):

$$ \sqrt{x} = -2 $$

Since the principal square root of a real number, \(\sqrt{x}\), cannot be negative, this equation has no real solution.

Step 4: Verify the solution

Let's check \( x = 256 \) in the original equation:

$$ 256 - 14\sqrt{256} - 32 = 0 $$

$$ 256 - 14(16) - 32 = 0 $$

$$ 256 - 224 - 32 = 0 $$

$$ 0 = 0 $$

The solution is valid.

Answer:

A. The solution set is \(\{256\}\).