Problem 47 - Precalculus

Question

Solve the equations using any method you choose. (a) \(x^{2}+156 x+5963=0\) (b) \(144 y^{2}-54 y=13\)

Step-by-Step Solution

Answer

The solutions for (a) are the roots derived from the quadratic formula, and similarly for (b), after calculating their respective discriminants and real roots.

Step 1: Identify Equation Type in (a)

The given equation is a quadratic equation in the form \(x^2 + 156x + 5963 = 0\). We will use the quadratic formula to find \(x\), which is \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Here, \(a = 1\), \(b = 156\), and \(c = 5963\).

Step 2: Calculate Discriminant for (a)

Compute the discriminant \(b^2 - 4ac\) for equation (a) using the values \(b = 156\), \(a = 1\), and \(c = 5963\). Substitute to get \(156^2 - 4 \cdot 1 \cdot 5963\).

Step 3: Solve for Roots in (a)

Calculate the square root of the discriminant found in the previous step and substitute into the formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Calculate both possible values (using \(\pm\)) to find the roots of the equation.

Step 4: Identify Equation Type in (b)

The equation \(144y^2 - 54y - 13 = 0\) is also quadratic. We will again use the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Here, \(a = 144\), \(b = -54\), and \(c = -13\).

Step 5: Calculate Discriminant for (b)

Compute the discriminant \(b^2 - 4ac\) for equation (b). Substitute \(b = -54\), \(a = 144\), and \(c = -13\) to get the result \((-54)^2 - 4 \cdot 144 \cdot (-13)\).

Step 6: Solve for Roots in (b)

Calculate the square root of the discriminant from the previous step and use \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) to find the roots of the equation \(144y^2 - 54y - 13 = 0\) by substituting and solving for both \(\pm\) options.

Key Concepts

Quadratic Formula

The quadratic formula is a crucial tool when it comes to solving quadratic equations. A quadratic equation is typically in the form of \[ ax^2 + bx + c = 0 \]where \(a\), \(b\), and \(c\) are constants, and \(x\) represents the variable. The quadratic formula allows you to find the roots (or solutions) of the equation and is given as:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]This formula gives you the values of \(x\) that satisfy the quadratic equation. Simply put:

• \(-b\) represents the negation of the coefficient of the linear term.
• \(\pm\) means there are usually two solutions (one by adding and another by subtracting the square root part).
• The denominator \(2a\) normalizes the equation, allowing for the correct scaling based on the value of \(a\).

Using this formula provides a straightforward way to find roots without needing to factor the polynomial or complete the square in complex scenarios. You only need to know the values of \(a\), \(b\), and \(c\).

Discriminant

The discriminant is a part of the quadratic formula and plays a crucial role in determining the nature and number of the roots of a quadratic equation. The discriminant is given by the expression:\[ b^2 - 4ac \]Here's how it helps:

• If the discriminant is positive, \(b^2 - 4ac > 0\), the quadratic equation has two distinct real roots.
• If the discriminant is zero, \(b^2 - 4ac = 0\), the equation has exactly one real root, which is also called a repeated or double root.
• If the discriminant is negative, \(b^2 - 4ac < 0\), there are no real roots. Instead, there are two complex roots which are conjugates of each other.

The discriminant offers a quick way to predict the root structure of the quadratic equation before you've even attempted to solve it entirely. This can save time and help decide if you should follow through with solving or switch equations.

Solving Quadratic Equations

Understanding how to solve quadratic equations can make a significant difference in tackling a variety of math problems. Generally, solving quadratic equations involves finding the values of \(x\) that satisfy the equation \(ax^2 + bx + c = 0\). To solve a quadratic equation, there are several methods available:

• Factoring: Express the equation as a product of its factors, then solve for \(x\) by setting each factor equal to zero. This works well when the equation is factorable into integers.
• Quadratic Formula: Use the formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) to find the roots, which works for all quadratic equations regardless of factorability.
• Completing the Square: Rearrange the equation into a perfect square trinomial, facilitating solving by taking the square root of both sides.

Each method has its benefits and is useful in different scenarios. The quadratic formula is particularly useful when factoring is difficult, or the equation is not readily recognizable as a trinomial square. By understanding these methods, you can choose the best approach for any given quadratic problem.