Problem 44 - Algebra and Trigonometry

Question

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

Step-by-Step Solution

Answer

The solutions are given by \(x = \frac{-2 \pm \sqrt{4 + 4\sqrt{3}}}{2\sqrt{6}}\).

Step 1: Identify the Equation

We are given the equation \(\sqrt{6} x^{2}+2x-\sqrt{\frac{3}{2}}=0\). This is a quadratic equation in the form \(ax^2 + bx + c = 0\).

Step 2: Define Coefficients

Identify the coefficients in the quadratic equation: \(a = \sqrt{6}\), \(b = 2\), and \(c = -\sqrt{\frac{3}{2}}\).

Step 3: Use the Quadratic Formula

Use the quadratic formula to find the solutions: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Here, substitute \(a = \sqrt{6}\), \(b = 2\), and \(c = -\sqrt{\frac{3}{2}}\).

Step 4: Calculate the Discriminant

Calculate the discriminant: \(b^2 - 4ac = 2^2 - 4 \cdot \sqrt{6} \cdot (-\sqrt{\frac{3}{2}})\). Simplify: \(4 + 4\sqrt{6} \cdot \sqrt{\frac{3}{2}}\).

Step 5: Simplify the Discriminant

Simplify further: \(4 + 4 \times \sqrt{3} = 4 + 4 \sqrt{3}\), using the fact that \(\sqrt{6} \cdot \sqrt{\frac{3}{2}} = \sqrt{9} = 3\).

Step 6: Solve for Roots

Substitute back into the quadratic formula: \(x = \frac{-2 \pm \sqrt{4 + 4\sqrt{3}}}{2\sqrt{6}}\). Simplify further to get possible roots.

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations like the one we were given: \(\sqrt{6}x^2 + 2x - \sqrt{\frac{3}{2}} = 0\). Quadratic equations take the form of \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants. In our example, these constants are:

• \(a = \sqrt{6}\)
• \(b = 2\)
• \(c = -\sqrt{\frac{3}{2}}\)

To find the solutions for \(x\), we plug these values into the quadratic formula:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This formula gives us the roots of the quadratic equation, meaning it tells us where the curve of the quadratic crosses the \(x\)-axis. Knowing how to apply this formula is essential for solving quadratic equations, especially complex ones involving square roots.

Discriminant in Quadratics

The discriminant is a component of the quadratic formula that provides insight into the nature of the roots without solving the entire equation. It is represented by the expression \(b^2 - 4ac\). For the given equation \(\sqrt{6}x^2 + 2x - \sqrt{\frac{3}{2}} = 0\), the calculations are as follows:

• Calculate \(b^2 = 2^2 = 4\)
• Calculate \(-4ac = -4 \cdot \sqrt{6} \cdot (-\sqrt{\frac{3}{2}}) = 4 \sqrt{6} \cdot \sqrt{\frac{3}{2}} = 4 \times 3 = 12\)

The discriminant is \(4 + 12 = 16\).When the discriminant is:

• Positive: there are two distinct real roots.
• Zero: there is exactly one root (a repeated root).
• Negative: there are no real roots, only complex ones.

In our example, a positive discriminant tells us there are two real solutions.

Real Solutions of Quadratic Equations

Real solutions of a quadratic equation are the values of \(x\) that satisfy the equation such that the equation equals zero. These solutions correspond to the points where the graph of the quadratic, a parabola, crosses the \(x\)-axis.In the given example \(\sqrt{6}x^2 + 2x - \sqrt{\frac{3}{2}} = 0\), we have previously found that the discriminant is positive. This indicates that the quadratic equation has two real roots. This is important because it tells us that the curve will intersect the \(x\)-axis at two distinct points, meaning the quadratic has two distinct real solutions.By substituting the values into the quadratic formula, we compute:\[x = \frac{-2 \pm \sqrt{4 + 4 \sqrt{3}}}{2\sqrt{6}}\]To obtain the exact real solutions, we'd further simplify the expression under the square root and complete the solution process. Identifying real solutions helps in graphing the equation and understanding the behavior of the quadratic in a visual context.