Question

Find the real solutions of each equation. Use a calculator to express any solutions rounded to two decimal places. $$ x-4 x^{1 / 2}+2=0 $$

Short answer

The real solutions are approximately 11.66 and 0.34.

Step-by-step solution

Substitute the square root

Let’s make the substitution: Let \(y = x^{\frac{1}{2}}\). This will transform the equation into a quadratic form. That gives us the new equation \(x = y^2\), so the equation becomes \(y^2 - 4y + 2 = 0\).

Apply the quadratic formula

The quadratic formula is given by \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). For our equation \(y^2 - 4y + 2 = 0\), we have \(a = 1\), \(b = -4\), and \(c = 2\). Substitute these values into the quadratic formula: \(y = \frac{4 \pm \sqrt{16 - 8}}{2}\).

Simplify the quadratic formula

Simplify the expression under the square root \(\sqrt{8}\) and then the entire formula: \(y = \frac{4 \pm \sqrt{8}}{2}\). Because \(\sqrt{8} = 2\sqrt{2}\), the expression simplifies to: \(y = \frac{4 \pm 2\sqrt{2}}{2} = 2 \pm \sqrt{2}\).

Substitute back for x

Recall that \(y = x^{\frac{1}{2}}\). Hence, \(x = y^2\). We have two solutions for \(y\): \(2 + \sqrt{2}\) and \(2 - \sqrt{2}\). Calculate \(x\) for each: \((2 + \sqrt{2})^2 = 4 + 4\sqrt{2} + 2 = 6 + 4\sqrt{2}\) and \((2 - \sqrt{2})^2 = 4 - 4\sqrt{2} + 2 = 6 - 4\sqrt{2}\).

Use a calculator and round

Find the decimal values of \(6 + 4\sqrt{2} \approx 11.66\) and \(6 - 4\sqrt{2} \approx 0.34\).

Key concepts

Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations of the form \(ax^2 + bx + c = 0\). Using the formula, you can find the roots of any quadratic equation. The formula is \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]. It involves substituting the coefficients \(a\), \(b\), and \(c\) into the formula and solving for \(x\). Let's break it down step-by-step:

• Identify the coefficients: \(a\), \(b\), and \(c\).
• Compute the value under the square root (discriminant) \(b^2 - 4ac\).
• Calculate the values of \(x\) using the formula.

The quadratic formula can be very handy, especially when factoring is difficult or impossible.

Real Solutions

In dealing with quadratic equations, you may encounter different types of solutions based on the discriminant (\(b^2 - 4ac\)). The solutions can be real or complex. Real solutions occur when the discriminant is non-negative (\(b^2 - 4ac \geq 0\)). When the discriminant is positive, you get two distinct real solutions:

• If the discriminant is zero, there is exactly one real solution, which is known as a repeated or double root.
• If the discriminant is negative, no real solutions exist; instead, you have complex solutions.

For the equation in our exercise, the discriminant \(16 - 8 = 8\) is positive, which means we have two distinct real solutions.

Substitution Method

When dealing with equations that are not easily recognizable as quadratic forms, the substitution method is very useful. The substitution method involves introducing a new variable to simplify the equation. For example, in our exercise, we used the substitution \(y = x^{1/2}\). This transforms the original equation into a quadratic one in terms of \(y\):

• First, solve for \(y\) using the quadratic formula.
• Once you have the solutions for \(y\), substitute back to find the values for the original variable \(x\).

This approach often simplifies the process and avoids more complex algebraic manipulations.

Square Roots

Working with square roots is crucial when solving quadratic equations, especially when simplifying the results from the quadratic formula. The square root function, \(\sqrt{x}\), gives us a number which, when squared, returns the original value.
Consider the simplification step in our exercise, where we encountered the expression \(\frac{4 \pm 2\sqrt{2}}{2}\). This reduces to \(2 \pm \sqrt{2}\). When dealing with roots, always remember:

• Simplify the expression wherever possible.
• Perform any necessary arithmetic operations under the root before division or multiplication.

Once simplified, substitute back into the original variable, as demonstrated. Working with square roots can seem tricky, but with practice, it becomes a straightforward process.