Problem 46 - Algebra for College Students

Question

Solve each quadratic equation using the method that seems most appropriate. $$(x-3)(x+5)=-7$$

Step-by-Step Solution

Answer

The solutions are \(x = 2\) and \(x = -4\).

Step 1: Expand the Quadratic Equation

First, we'll expand the given equation to express it in standard quadratic form. We start by using the distributive property on the left side: \((x-3)(x+5) = x^2 + 5x - 3x - 15\). Combine like terms to get: \(x^2 + 2x - 15\). Thus, the equation becomes: \(x^2 + 2x - 15 = -7\).

Step 2: Bring all Terms to One Side

Add 7 to both sides to set the equation to zero:\(x^2 + 2x - 15 + 7 = 0\).Simplify to get:\(x^2 + 2x - 8 = 0\). Now, the equation is ready to be solved using the quadratic formula.

Step 3: Apply the Quadratic Formula

The quadratic formula is given by:\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where for the equation \(x^2 + 2x - 8 = 0\), we have \(a = 1\), \(b = 2\), and \(c = -8\).Substitute these values into the formula:\(x = \frac{-2 \pm \sqrt{2^2 - 4 \times 1 \times (-8)}}{2 \times 1} = \frac{-2 \pm \sqrt{4 + 32}}{2}\).

Step 4: Calculate the Discriminant

Calculate the discriminant inside the square root:\(2^2 - 4(-8) = 4 + 32 = 36\).This discriminant is positive, indicating two real solutions.

Step 5: Find the Two Solutions

Substitute the discriminant back into the quadratic formula:\(x = \frac{-2 \pm \sqrt{36}}{2}\).Calculate the two possible solutions:\(x = \frac{-2 + 6}{2} = 2\) and \(x = \frac{-2 - 6}{2} = -4\).Thus, the solutions are \(x = 2\) and \(x = -4\).

Key Concepts

Quadratic Formula

One of the most powerful tools for solving quadratic equations is the quadratic formula. It is given by \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). This formula allows us to find the solutions for any quadratic equation of the form \( ax^2 + bx + c = 0 \). Here, \( a \), \( b \), and \( c \) are coefficients from the equation, and the solutions for \( x \) are obtained by plugging these values into the formula.

• "\(-b\)" refers to the opposite of the coefficient of \( x \).
• "\(b^2 - 4ac\)" is known as the discriminant, providing key information about the nature of the solutions.
• The "\(\pm\)" sign indicates that there can be two solutions for \( x \), due to the addition or subtraction from the square root.
• The whole expression is divided by "\(2a\)" to finalize the values of \( x \).

Understanding how to use the quadratic formula is crucial, as it provides an efficient and direct approach to finding solutions for quadratic equations when factoring or other methods are not suitable.

Discriminant

The discriminant is an integral part of the quadratic formula and is represented by the expression \( b^2 - 4ac \). It helps determine the nature of the solutions for a quadratic equation without actually solving it. The value of the discriminant can tell us the following:

• If \( b^2 - 4ac > 0 \), there are two distinct real solutions, meaning the parabola crosses the x-axis at two points.
• If \( b^2 - 4ac = 0 \), there is exactly one real solution, meaning the parabola just touches the x-axis at one point, called a double root or repeated root.
• If \( b^2 - 4ac < 0 \), there are no real solutions, meaning the parabola does not intersect the x-axis at all and instead sits fully above or below it.

This little check can save time and guide your problem-solving strategy by predicting how many real solutions you can expect.

Standard Form of Quadratic Equation

The standard form of a quadratic equation is \( ax^2 + bx + c = 0 \). It provides a clear framework for identifying the coefficients needed when using the quadratic formula.

• "\(a\)" is the coefficient of \(x^2\), which dictates how "wide" or "narrow" the parabola opens.
• "\(b\)" is the coefficient of \(x\), affecting the direction and position of the parabola's axis of symmetry.
• "\(c\)" is the constant term and represents the y-intercept, where the parabola crosses the y-axis.

To utilize methods like the quadratic formula, rewriting equations into this standard form is often required. Ensuring all terms are moved to one side of the equation to equal zero is an essential step to solve the equation correctly.

Real Solutions

Real solutions of a quadratic equation are the values of \( x \) where the equation equals zero, leading to intersections of the parabola with the x-axis. For a quadratic equation to have real solutions, the discriminant must be greater than or equal to zero.

• If the discriminant is positive, we get two real solutions. These solutions represent the x-coordinates where the graph of the quadratic equation crosses the x-axis.
• If the discriminant is zero, there is exactly one real solution, indicating that the parabola touches the x-axis at one point, meaning the vertex is on the x-axis.
• In cases where the discriminant is negative, the quadratic equation has no real solutions because the parabola does not intersect the x-axis.

Understanding these solutions helps in graphically interpreting quadratic equations, and knowing about them aids in anticipating and verifying calculations.