Problem 35 - Algebra and Trigonometry

Question

Find all real solutions of the equation. \(3 x^{2}+6 x-5=0\)

Step-by-Step Solution

Answer

The real solutions are \(x = -1 + \frac{2\sqrt{6}}{3}\) and \(x = -1 - \frac{2\sqrt{6}}{3}\).

Step 1: Identify the Quadratic Equation

The given equation is a quadratic equation of the standard form, \(ax^2 + bx + c = 0\), where \(a = 3\), \(b = 6\), and \(c = -5\).

Step 2: Calculate the Discriminant

The discriminant \(D\) of a quadratic equation \(ax^2 + bx + c = 0\) is given by the formula \(D = b^2 - 4ac\). Substitute the values to find \(D = 6^2 - 4 \times 3 \times (-5) = 36 + 60 = 96\).

Step 3: Check the Nature of the Roots

Since the discriminant \(D = 96\) is greater than zero, the quadratic equation has two distinct real solutions.

Step 4: Use the Quadratic Formula

The solutions of the quadratic equation are given by the quadratic formula: \(x = \frac{-b \pm \sqrt{D}}{2a}\). Substitute the values we have: \(x = \frac{-6 \pm \sqrt{96}}{6}\).

Step 5: Simplify the Square Root

First simplify \(\sqrt{96}\). Since \(96 = 16 \times 6\), we can write \(\sqrt{96} = \sqrt{16 \times 6} = \sqrt{16} \times \sqrt{6} = 4\sqrt{6}\).

Step 6: Solve for the Roots

Plug \(\sqrt{96} = 4\sqrt{6}\) into the quadratic formula: \(x = \frac{-6 \pm 4\sqrt{6}}{6}\). Simplify to get the solutions: \(x = \frac{-6 + 4\sqrt{6}}{6}\) and \(x = \frac{-6 - 4\sqrt{6}}{6}\). Further simplification gives \(x = -1 + \frac{2\sqrt{6}}{3}\) and \(x = -1 - \frac{2\sqrt{6}}{3}\).

Key Concepts

Discriminant

The discriminant is a key component when solving quadratic equations, as it helps determine the nature of the roots. A quadratic equation takes the form \(ax^2 + bx + c = 0\). The discriminant \(D\) for this equation is calculated as \(D = b^2 - 4ac\).

The value of the discriminant tells us about the potential solutions:

• If \(D > 0\), there are two distinct real roots.
• If \(D = 0\), there is exactly one real root, sometimes referred to as a double root.
• If \(D < 0\), there are no real roots, but two complex roots instead.

In our exercise, we calculated \(D = 96\), which is greater than 0, indicating that the quadratic equation has two distinct real solutions.

Solutions of a Quadratic Equation

Finding the solutions of a quadratic equation means identifying the values of \(x\) that satisfy the equation \(ax^2 + bx + c = 0\).

Once you compute the discriminant, it guides you on how the solutions will look:

• With two distinct real solutions (when \(D > 0\)), you know both solutions will be different real numbers.
• If there is exactly one real solution (\(D = 0\)), the equation results in a perfect square trinomial.
• For no real solutions (\(D < 0\)), solutions turn out to be complex numbers.

Using the information from the discriminant, you can then proceed to find these solutions using the quadratic formula or by factoring.

Quadratic Formula

The quadratic formula is a trustworthy method to find the solutions of any quadratic equation of the form \(ax^2 + bx + c = 0\).

The formula is given by:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This clever formula gives both potential solutions at once:

• \(\frac{-b + \sqrt{b^2 - 4ac}}{2a}\)
• \(\frac{-b - \sqrt{b^2 - 4ac}}{2a}\)

You substitute the values of \(a, b,\) and \(c\) from the given equation into this formula. In our example, substituting these values yielded \(x = -1 + \frac{2\sqrt{6}}{3}\) and \(x = -1 - \frac{2\sqrt{6}}{3}\).

This solution approach is particularly useful when factoring is difficult or impossible, reinforcing why the quadratic formula is an essential tool in algebra.