Problem 49 - Algebra and Trigonometry

Question

Find all real solutions of the equation. $$ 10 y^{2}-16 y+5=0 $$

Step-by-Step Solution

Answer

The solutions are \( y = \frac{8 + \sqrt{14}}{10} \) and \( y = \frac{8 - \sqrt{14}}{10} \).

Step 1: Identify the quadratic equation coefficients

The given equation is in the form of a standard quadratic equation: \( ax^2 + bx + c = 0 \). Here, \( a = 10 \), \( b = -16 \), and \( c = 5 \).

Step 2: Use the quadratic formula

The quadratic formula is \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Substitute \( a = 10 \), \( b = -16 \), and \( c = 5 \) into this formula to find the solutions for \( y \).

Step 3: Calculate the discriminant

First, calculate the discriminant \( \Delta = b^2 - 4ac \). Substitute the values to find \( \Delta = (-16)^2 - 4 \cdot 10 \cdot 5 = 256 - 200 = 56 \). Since the discriminant is positive, there are two real solutions.

Step 4: Calculate the square root of the discriminant

Find \( \sqrt{56} \). Simplify this to \( \sqrt{56} = \sqrt{4 \cdot 14} = 2 \sqrt{14} \).

Step 5: Substitute into the quadratic formula

Substitute \( b = -16 \), \( \sqrt{\Delta} = 2\sqrt{14} \), and \( a = 10 \) into the quadratic formula: \[ y = \frac{-(-16) \pm 2 \sqrt{14}}{2 \cdot 10} = \frac{16 \pm 2 \sqrt{14}}{20} \].

Step 6: Simplify the expression

Simplify the expression by dividing all terms by 2: \[ y = \frac{16 \pm 2 \sqrt{14}}{20} = \frac{8 \pm \sqrt{14}}{10} \].Thus, the solutions are \( y = \frac{8 + \sqrt{14}}{10} \) and \( y = \frac{8 - \sqrt{14}}{10} \).

Key Concepts

Quadratic Formula

Quadratic equations are expressions where the variable is raised to the power of two. In simpler terms, they have the form: \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). The quadratic formula is a powerful tool for finding the values of \( x \) that make the equation true. This formula is:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

It gives you two solutions, hence the plus or minus (\( \pm \)) sign. These solutions are essentially the x-values where the parabola intersects the x-axis. By plugging in your specific values for \( a \), \( b \), and \( c \), you can calculate these intersection points. To solve our example equation, we identified our coefficients as \( a = 10 \), \( b = -16 \), and \( c = 5 \). Next, we substitute these values into the quadratic formula to find our solutions.

Discriminant

The discriminant is a part of the quadratic formula, which tells us about the nature of the roots of a quadratic equation. It's calculated using the expression \( \Delta = b^2 - 4ac \).

• If \( \Delta > 0 \), the quadratic equation has two distinct real solutions.
• If \( \Delta = 0 \), there is exactly one real solution.
• If \( \Delta < 0 \), there are no real solutions; instead, the solutions are complex.

For our equation, substituting \( b = -16 \), \( a = 10 \), and \( c = 5 \) gives \( \Delta = 256 - 200 = 56 \). Since \( \Delta \) (56) is positive, we know there are two separate real solutions. This means the graph of the equation crosses the x-axis at two points.

Real Solutions

When solving a quadratic equation, finding the real solutions means determining the specific x-values that satisfy the equation. Real solutions are the points where the graph of the quadratic function hits the x-axis. In this case, after using the quadratic formula, our solutions are:

• \( y = \frac{8 + \sqrt{14}}{10} \)
• \( y = \frac{8 - \sqrt{14}}{10} \)

These are our two distinct real solutions for the quadratic equation \( 10y^2 - 16y + 5 = 0 \). Remember, real solutions occur when the discriminant is positive, indicating the parabola touches the x-axis at two points. The solutions indicate exactly where on the x-axis these points of intersection occur.