Problem 28 - Basic Technical Mathematics with Calculus

Question

Solve the given quadratic equations by factoring. $$t(43+t)=9-9 t^{2}$$

Step-by-Step Solution

Answer

The solutions are \(t = \frac{1}{5}\) and \(t = -\frac{9}{2}\).

Step 1: Expand and Rearrange the Equation

To solve the quadratic equation, begin by expanding the left-hand side and rearranging all terms to one side. Given the equation:\[t(43 + t) = 9 - 9t^2\]First, distribute the \(t\) on the left side:\[43t + t^2 = 9 - 9t^2\]Now, move all terms to one side of the equation to set it to zero:\[43t + t^2 + 9t^2 - 9 = 0\]Combine like terms:\[t^2 + 9t^2 + 43t - 9 = 0\]Simplify the equation:\[10t^2 + 43t - 9 = 0\]This is the quadratic equation we will factor.

Step 2: Factor the Quadratic Equation

The quadratic equation we need to factor is:\[10t^2 + 43t - 9 = 0\]To factor this, search for two numbers whose product is \(10 imes (-9) = -90\) and whose sum is \(43\).We find that \(45\) and \(-2\) satisfy these conditions, as:\[45 imes (-2) = -90\]\[45 + (-2) = 43\]Now, rewrite the middle term using these two numbers:\[10t^2 + 45t - 2t - 9 = 0\]Group the terms for factoring by grouping:\[(10t^2 + 45t) - (2t + 9) = 0\]Factor each group:\[5t(2t + 9) - 1(2t + 9) = 0\]Factor out the common binomial factor:\[(5t - 1)(2t + 9) = 0\]

Step 3: Solve for the Roots

Now that the equation is factored as:\[(5t - 1)(2t + 9) = 0\]Set each factor equal to zero and solve for \(t\):1. \(5t - 1 = 0\) Solving for \(t\): \[ 5t = 1 \ t = \frac{1}{5} \]2. \(2t + 9 = 0\) Solving for \(t\): \[ 2t = -9 \ t = -\frac{9}{2} \]The solutions for \(t\) are \(t = \frac{1}{5}\) and \(t = -\frac{9}{2}\).

Key Concepts

Factoring quadratic equations

Factoring quadratic equations is a powerful technique often used to find the roots of the equation. It involves expressing the quadratic equation in the form of two binomial expressions set to zero, making it easier to identify the solutions. For the equation presented in the exercise \[10t^2 + 43t - 9 = 0\]we can factor it into \[(5t - 1)(2t + 9) = 0\]. This is accomplished through a process called factoring by grouping.To begin factoring, start by looking for two numbers whose product equals the coefficient of the quadratic term times the constant term (in this case \(10 \times -9 = -90\)) and whose sum equals the linear coefficient (43 in our equation). Once these numbers are found, as in 45 and -2, the linear term (intermediate term) is rewritten using these numbers.

• First, break the middle term into two terms, \(45t\) and \(-2t\).
• Then, group the terms into pairs and factor them separately.
• Finally, factor out the common binomial factor in each group.

Factoring transforms the quadratic equation into a simpler form, allowing you to solve for the roots easily.

Roots of quadratic equation

The roots of a quadratic equation are the solutions to the equation. In essence, they are the values of the variable that make the equation true. Once the quadratic equation is factored into the form \(ax + b)(cx + d) = 0\), finding the roots becomes straightforward.For the factored equation \(5t - 1)(2t + 9) = 0\), you find the roots by setting each binomial expression equal to zero and solving for the variable. This gives you two separate equations:

• \(5t - 1 = 0\)
• \(2t + 9 = 0\)

Solving these will yield the values of \(t\) that satisfy the equation:

• From \(5t - 1 = 0\), solving for \(t\) gives \(t = \frac{1}{5}\).
• From \(2t + 9 = 0\), solving for \(t\) gives \(t = -\frac{9}{2}\).

These roots represent the points where the graph of the quadratic touches or crosses the \(t\)-axis.

Equation expansion and simplification

Expanding and simplification are crucial initial steps when working with quadratic equations. These steps help in transforming the given equation into a standard form, which is essential for applying factorization techniques.Starting from the equation \t(43 + t) = 9 - 9t^2\, our first step is to expand. This involves distributing the multiplication across the terms inside the parentheses, resulting in \43t + t^2\. Rearranging involves moving all terms to one side:

• Add \(9t^2\) to \(t^2\), resulting in \(10t^2\).
• Subtract 9 from the constant terms, aligning everything to equal zero.

Thus, we have the expression \(10t^2 + 43t - 9 = 0\), a simplified quadratic that can be factored. Simplification involves combining like terms while ensuring you have the equation in a recognizable quadratic form \(ax^2 + bx + c = 0\). This standard form is essential as it helps us apply the factoring method effectively to solve for the roots of the quadratic equation.