Problem 33 - Algebra and Trigonometry

Question

Find all real solutions of the equation. $$ 2 x^{2}+x-3=0 $$

Step-by-Step Solution

Answer

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

Step 1: Identify the Quadratic Equation Coefficients

The given quadratic equation is \(2x^2 + x - 3 = 0\). Identify the coefficients: \(a = 2\), \(b = 1\), and \(c = -3\).

Step 2: Calculate the Discriminant

Use the discriminant formula \(b^2 - 4ac\). Substitute \(a = 2\), \(b = 1\), and \(c = -3\) into the formula: \((1)^2 - 4(2)(-3) = 1 + 24 = 25\). The discriminant is positive, indicating there are two real solutions.

Step 3: Use the Quadratic Formula

The quadratic formula is \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Substitute \(a = 2\), \(b = 1\), and \(c = -3\), as well as \(b^2 - 4ac = 25\), into the formula: \(x = \frac{-1 \pm \sqrt{25}}{4}\).

Step 4: Compute the Solutions

Calculate the two solutions using the quadratic formula: 1. \(x_1 = \frac{-1 + 5}{4} = 1\)2. \(x_2 = \frac{-1 - 5}{4} = -\frac{3}{2}\).

Step 5: Verify the Solutions

To verify, substitute \(x = 1\) and \(x = -\frac{3}{2}\) back into the original equation:\(2(1)^2 + 1 - 3 = 0\) and \(2((-\frac{3}{2})^2) + (-\frac{3}{2}) - 3 = 0\). Both satisfy the equation.

Key Concepts

Discriminant

When solving quadratic equations, the discriminant is a key component. It is part of the quadratic formula and helps us determine the nature of the solutions we can expect. In a quadratic equation of the form \(ax^2 + bx + c = 0\), the discriminant \(D\) is calculated using the formula:\[D = b^2 - 4ac\]The value of the discriminant tells us the nature of the roots of the quadratic equation:

• If \(D > 0\), the quadratic equation has two distinct real solutions.
• If \(D = 0\), there is exactly one real solution or a repeated real root.
• If \(D < 0\), the equation has no real solutions, but rather two complex solutions.

In our exercise, we found that the discriminant is 25, which is greater than zero. Therefore, the given quadratic equation has two distinct real solutions.

Quadratic Formula

The quadratic formula is a powerful tool for finding the solutions of any quadratic equation. It can be used regardless of the complexity of the coefficients. The formula for solving the quadratic equation \(ax^2 + bx + c = 0\) is:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This formula provides a straightforward method for calculating the roots of the equation by substituting the values of \(a\), \(b\), and \(c\). In the original exercise, we identified the coefficients as \(a = 2\), \(b = 1\), and \(c = -3\). By substituting these values into the formula along with the discriminant \(b^2 - 4ac = 25\), the formula becomes:\[x = \frac{-1 \pm 5}{4}\]This results in two possible solutions for \(x\), which we compute to find the real solutions of the initial equation.

Real Solutions

Real solutions are the values of \(x\) that satisfy the given quadratic equation. After applying the quadratic formula, we calculated the solutions for our specific equation as \(x = 1\) and \(x = -\frac{3}{2}\). These are the real solutions we were able to find:

• \(x_1 = \frac{-1 + 5}{4} = 1\)
• \(x_2 = \frac{-1 - 5}{4} = -\frac{3}{2}\)

To ensure these solutions are correct, we substitute them back into the original equation:

• For \(x = 1\): \(2(1)^2 + 1 - 3 = 0\), which simplifies to \(0\).
• For \(x = -\frac{3}{2}\): \(2(-\frac{3}{2})^2 + (-\frac{3}{2}) - 3 = 0\), which also simplifies to \(0\).

Both substitutions prove that \(x = 1\) and \(x = -\frac{3}{2}\) are indeed the correct real solutions.