Problem 45 - 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 \(y = \frac{4 \pm \sqrt{14}}{5}\).

Step 1: Identify the Equation Form

Recognize that the equation is a quadratic of the form \(ay^2 + by + c = 0\). For the given equation, \(a = 10\), \(b = -16\), and \(c = 5\).

Step 2: Apply the Quadratic Formula

Use the quadratic formula to find the solutions of the quadratic equation: \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).

Step 3: Calculate the Discriminant

Compute the discriminant \(b^2 - 4ac\): \((-16)^2 - 4(10)(5) = 256 - 200 = 56\).

Step 4: Evaluate the Square Root of the Discriminant

Calculate \(\sqrt{56}\). This simplifies to \(\sqrt{4 \times 14} = \sqrt{4} \times \sqrt{14} = 2\sqrt{14}\).

Step 5: Substitute in the Quadratic Formula

Insert the calculated values into the quadratic formula: \(y = \frac{-(-16) \pm 2\sqrt{14}}{2 \times 10}\).

Step 6: Simplify the Solutions

Simplify the expression: \(y = \frac{16 \pm 2\sqrt{14}}{20} = \frac{16}{20} \pm \frac{2\sqrt{14}}{20}\), which results in \(y = \frac{4}{5} \pm \frac{\sqrt{14}}{10}\).

Step 7: State the Real Solutions

Thus, the real solutions are \(y = \frac{4 + \sqrt{14}}{5}\) and \(y = \frac{4 - \sqrt{14}}{5}\).

Key Concepts

Quadratic Formula

The quadratic formula is an essential tool for solving quadratic equations, which are equations of the form \( ax^2 + bx + c = 0 \). This formula provides a straightforward method to find the solutions (roots) of any quadratic equation. The quadratic formula is given by:

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

The variables \( a \), \( b \), and \( c \) represent coefficients from the quadratic equation, where \( a eq 0 \). By substituting the coefficients of your specific equation into the formula, you can calculate the solutions. It is incredibly helpful because it works no matter what values \( a \), \( b \), and \( c \) are, as long as \( a eq 0 \). You only need to know these coefficients to proceed with finding solutions using the quadratic formula.

Discriminant

The discriminant is a significant component of the quadratic formula. It is denoted by \( b^2 - 4ac \) and it is located under the square root symbol in the quadratic formula:

• \( \Delta = b^2 - 4ac \)

The discriminant helps determine the nature of the roots for the quadratic equation:

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

Understanding the discriminant is crucial as it tells you how many real solutions are possible even before you go on to solve the equation itself.

Real Solutions

Real solutions are the results you get when you solve a quadratic equation and find that the discriminant is non-negative. This means you either have:

• Two distinct real solutions if the discriminant is greater than zero.
• One real solution if the discriminant equals zero.

In our problem, after calculating the discriminant and finding that it is 56, which is greater than zero, we know to expect two real solutions. These solutions represent the x-values where the quadratic equation equals zero, graphically shown as the x-intercepts of the quadratic function.

Simplifying Radicals

Simplifying radicals is an algebraic process used to make expressions involving square roots (and other roots) more manageable and easier to understand. The process involves finding and factoring out perfect squares from under the radical and simplifying them:

• For example, \( \sqrt{56} \) is the same as \( \sqrt{4 \times 14} \).
• Since \( \sqrt{4} \) is a perfect square and equals 2, \( \sqrt{4 \times 14} \) can be simplified to \( 2\sqrt{14} \).

In our quadratic equation solution, simplifying \( \sqrt{56} \) to \( 2\sqrt{14} \) made the rest of our calculations more straightforward, leading us to simpler expressions and ultimately, the final real solutions of the equation.