Problem 17 - Algebra and Trigonometry Real Mathematics, Real People

Question

Solve the quadratic equation by factoring. Check your solutions in the original equation. $$-x^{2}-11 x=28$$

Step-by-Step Solution

Answer

The correct solution to the equation is x = -4.

Step 1: Rearrange the equation

The given equation is \( -x^{2} - 11x = 28 \). Let's rearrange the equation to look like this: \( -x^{2} - 11x - 28 = 0 \)

Step 2: Factor the equation

We are going to factor the equation now. To do that, we need two numbers that add up to -11 and multiply to -28. These numbers are -4 and 7. Hence you can write the equation as \( -x^{2} + 7x - 4x - 28 = 0 \) or \( -x(x - 7) -4(x - 7) = 0 \). Therefore, our factored equation is \( (x - 7) (-x -4)=0 \)

Step 3: Solving for x

Now, we will solve for x. We have two factors. If either one is equal to zero, the equation is true. So, we will set each one to zero and solve for x: \( x - 7=0 \rightarrow x=7 \), and \( -x -4=0 \rightarrow x=-4 \). Hence, we have found the solutions x = 7 and x = -4.

Step 4: Check solutions in the original equation

To check if the solutions are correct, we will substitute x with 7 and -4 in the original equation and simplify. If both simplified expressions equal 28, the solutions are correct. For x = 7: \( -7^{2} - 11*7 \rightarrow -49 - 77 = -126 \). That is not equal to 28. So, this is not a root of the equation. For x = -4: \( -(-4)^{2} - 11*-4 \rightarrow -16 + 44 = 28 \). That is equal to 28. So, x = -4 is a root of the equation. Therefore, the only solution for the equation is x = -4.

Key Concepts

Factoring Quadratic Equations

When we tackle quadratic equations, one of the simplest methods we can use is factoring. This techniques hinges on rewriting the quadratic equation in a format where it is the product of two binomials or a monomial and a binomial. Essentially, we are looking for two numbers that when multiplied together give the product of the quadratic coefficient and the constant term, and when added together, provide the coefficient of the linear term.

For the equation \( -x^2 - 11x - 28 = 0 \), we reshuffle it slightly to make it easier to factor, as demonstrated in the steps. The goal is to find a combination of numbers that multiply to give -28 and add to give -11. In this case, -4 and 7 work perfectly. After discovering these numbers, breaking down the equation into its factored components becomes straightforward: \((x-7)(-x-4)=0\). Factoring not only simplifies the problem but sets the stage for finding the roots of the equation through simple algebraic manipulation.

Quadratic Formula

When factoring is difficult or impossible, the quadratic formula is a reliable tool. It is a standard formula derived from the equation of a quadratic function, and it can solve any quadratic equation. The formula is \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a\), \(b\), and \(c\) represent the coefficients of the terms in the standard form of the quadratic equation \(ax^2 + bx + c = 0\).

The discriminant, \(b^2 - 4ac\), tells us about the nature of the roots. A positive discriminant implies two real and distinct solutions, zero means one real solution, and a negative one indicates complex solutions. Even though the initial problem was solved by factoring, it's important to acknowledge how useful the quadratic formula is, especially when the roots are not easily guessable.

Roots of Quadratic Equations

The roots of a quadratic equation are the values of \(x\) that make the equation true when substituted back into the equation. These roots can be real or complex numbers, depending on the equation's coefficients. When we factored the example \( (x-7)(-x-4) = 0 \), we are essentially looking for the roots of the equation.

By setting each factor equal to zero, we can solve for the roots: \((x-7)=0\) and \((-x-4)=0\). The solutions to these individual equations, \(x=7\) and \(x=-4\), are the candidates for the roots of the original equation. It's crucial to note that not every solution derived from the factored form might satisfy the original equation, which is why checking these solutions is a critical step.

Check Solutions

Checking your solutions is vital in the problem-solving process. It verifies the accuracy of the discovered roots and ensures they satisfy the original equation. In the example provided, the potential solutions \(x=7\) and \(x=-4\) need to be checked by substituting them back into the original equation \( -x^2 - 11x = 28 \).

As seen, when \(x=7\), the equation does not hold true, while for \(x=-4\), the original equation is satisfied. Hence, \(x=-4\) is the correct solution. This checking step prevents the acceptance of false roots and confirms our final answer, providing a clear understanding of how roots function within quadratic equations.