Problem 58 - Algebra and Trigonometry

Question

Use the quadratic formula and a calculator to find all real solutions, rounded to three decimals. $$ 12.714 x^{2}+7.103 x=0.987 $$

Step-by-Step Solution

Answer

The real solutions are approximately 0.115 and -0.674.

Step 1: Identify the coefficients

Identify the coefficients from the quadratic equation given: \(ax^2 + bx + c = 0\). In this case, \(a = 12.714\), \(b = 7.103\), and \(c = -0.987\). Note that we bring all terms to one side so the equation becomes: \(12.714x^2 + 7.103x - 0.987 = 0\).

Step 2: Write the quadratic formula

The quadratic formula is given by: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). This formula allows us to find the roots of a quadratic equation.

Step 3: Calculate the discriminant

Calculate the discriminant \(\Delta\) using: \(\Delta = b^2 - 4ac\). Substitute the coefficients: \(\Delta = (7.103)^2 - 4(12.714)(-0.987)\).

Step 4: Evaluate the discriminant

Compute \(\Delta = 50.422209 - (-50.123912) = 50.422209 + 50.123912 = 100.546121\). Since the discriminant is positive, there are two real solutions.

Step 5: Apply the quadratic formula

Substitute \(a\), \(b\), and \(\Delta\) into the quadratic formula. Compute both solutions: \[x = \frac{-7.103 \pm \sqrt{100.546121}}{2 \cdot 12.714}\].

Step 6: Calculate the solutions

First compute \( \sqrt{100.546121} \approx 10.025\). Then compute each solution: \(x_1 = \frac{-7.103 + 10.025}{2 \times 12.714} = \frac{2.922}{25.428} \approx 0.115\) and \(x_2 = \frac{-7.103 - 10.025}{2 \times 12.714} = \frac{-17.128}{25.428} \approx -0.674\).

Step 7: Round to three decimals

The final solutions, rounded to three decimal places, are \(x_1 \approx 0.115\) and \(x_2 \approx -0.674\).

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool used to find the roots of any quadratic equation, which is an equation in the form of \( ax^2 + bx + c = 0 \). This formula provides us with a way to find solutions even when factoring is complex or impossible. It looks like this:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here's how it works:

• "\(-b\)" indicates taking the opposite sign of the coefficient in front of \( x \).
• The "\(+\)" and "\(-\)" signs mean you will calculate two different values, leading to two possible solutions for \(x\).
• The term under the square root, "\(b^2 - 4ac\)", is crucial and has a special name: the discriminant.
• Finally, you divide everything by \(2a\) to solve for \(x\).

Remember, using the quadratic formula involves straightforward substitution of the coefficients into the formula, allowing us to compute the real solutions efficiently. This method is reliable and exact when followed through carefully.

Discriminant

The discriminant is a key element in understanding the nature of the solutions of a quadratic equation. It is represented by the term \(b^2 - 4ac\) in the quadratic formula. The value of the discriminant tells us how many real solutions the quadratic equation has.

• If the discriminant is positive (\(\Delta > 0\)), there are two distinct real solutions. This is because the square root of a positive number results in two different values when you add and subtract them (hence the \(\pm \) in the formula).
• If the discriminant is zero (\(\Delta = 0\)), there is exactly one real solution because adding and subtracting zero gives the same result.
• If the discriminant is negative (\(\Delta < 0\)), there are no real solutions. Instead, the solutions are complex numbers, which means they include imaginary numbers.

In our example, the discriminant calculation resulted in a positive \(100.546121\), allowing us to expect two real solutions. It's an initial check to predict how many solutions we should find.

Real Solutions

Real solutions refer to the actual, tangible solutions to a quadratic equation that we can calculate using real numbers. When solving a quadratic equation, the goal is often to find these real solutions.

• We determine real solutions by applying the quadratic formula, only valid when the discriminant is non-negative (zero or positive).
• The plus or minus sign (\(\pm\)) within the formula gives two possibilities, hence potentially two real solutions.
• Each calculation provides a different solution for \(x\), which we found as \(x_1 = 0.115\) and \(x_2 = -0.674\) in the exercise.
• These solutions can be validated by plugging them back into the original equation to see if they satisfy the equation \(12.714x^2 + 7.103x = 0.987\).

Real solutions are the endpoints of successfully applying the quadratic formula, reflecting how the equation behaves graphically, often seen as intercepts on a coordinate plane graph.