Problem 37 - Algebra and Trigonometry

Question

Find all real solutions of the equation. \(2 y^{2}-y-\frac{1}{2}=0\)

Step-by-Step Solution

Answer

The real solutions are \(y = \frac{1 + \sqrt{5}}{4}\) and \(y = \frac{1 - \sqrt{5}}{4}\).

Step 1: Identify and Rearrange the Equation

The given equation is \(2y^2 - y - \frac{1}{2} = 0\). It is a quadratic equation in the form \(ax^2 + bx + c = 0\), where \(a = 2\), \(b = -1\), and \(c = -\frac{1}{2}\).

Step 2: Use the Quadratic Formula

The quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) can be used to find the roots. Here, substitute \(a = 2\), \(b = -1\), and \(c = -\frac{1}{2}\) into the formula.

Step 3: Calculate the Discriminant

Calculate the discriminant \(b^2 - 4ac\) which is \((-1)^2 - 4(2)(-\frac{1}{2})\). This simplifies to \(1 + 4 = 5\).

Step 4: Substitute Values into the Quadratic Formula

Substitute the values into the quadratic formula: \(y = \frac{-(-1) \pm \sqrt{5}}{2(2)} = \frac{1 \pm \sqrt{5}}{4}\).

Step 5: Solve for Both Solutions

Calculate both solutions. For the positive square root: \(y = \frac{1 + \sqrt{5}}{4}\) and for the negative square root: \(y = \frac{1 - \sqrt{5}}{4}\).

Step 6: Simplify the Solutions if Possible

Both solutions \(y = \frac{1 + \sqrt{5}}{4}\) and \(y = \frac{1 - \sqrt{5}}{4}\) are already in their simplest form.

Key Concepts

Quadratic Formula

The quadratic formula is an essential tool when dealing with quadratic equations, which are equations in the form of \( ax^2 + bx + c = 0 \). This formula helps us find the roots, or solutions, of these equations.
The formula is given by

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

Here, \( a \), \( b \), and \( c \) represent the coefficients from the quadratic equation. The symbol \( \pm \) means that you will end up with two possible solutions, one with a plus and one with a minus.
In our exercise, once we identified \( a = 2 \), \( b = -1 \), and \( c = -\frac{1}{2} \), we substituted these into the formula. This simplifies the process, allowing us to directly calculate the roots.

Discriminant Calculation

The discriminant is a part of the quadratic formula under the square root sign, represented by \( b^2 - 4ac \). It helps determine the nature of the roots of the quadratic equation.

• If the discriminant is positive, the equation has two distinct real roots.
• If it is zero, the equation has exactly one real root or a repeated real root.
• If the discriminant is negative, the equation has no real roots and instead has two complex roots.

In our example, we calculated the discriminant to be \( 5 \), which is positive. This indicates that there are two different real roots for our equation. Knowing the discriminant ahead of time gives us valuable information about how many solutions we should expect.

Roots of Equations

After using the quadratic formula and calculating the discriminant, the next step is to find the roots of the quadratic equation. Roots are simply the values of \( y \) that satisfy the equation.
Substituting the coefficients and the discriminant back into the quadratic formula, we get

• \( y = \frac{1 + \sqrt{5}}{4} \)
• \( y = \frac{1 - \sqrt{5}}{4} \)

These are the two roots or solutions of our quadratic equation \( 2y^2 - y - \frac{1}{2} = 0 \).
Each root represents a possible solution where the quadratic equation equals zero. In practical situations, these roots can correspond to points of intersection on a graph or values that satisfy a specific problem condition.