Problem 48 - Elementary Algebra

Question

Choose the appropriate method to solve the following. $$ 5 t(5 t-6)=-9 $$

Step-by-Step Solution

Answer

The solution is \( t = \frac{3}{5} \).

Step 1: Expand the Equation

First, expand the equation by applying the distributive property to clear the expression within the parentheses, i.e., multiply each term inside the parentheses by the term outside: \[ 5t(5t - 6) = 25t^2 - 30t \]Rewrite the equation: \[ 25t^2 - 30t = -9 \]

Step 2: Rearrange to Form a Quadratic Equation

To form a standard quadratic equation, move all terms to one side of the equation:\[ 25t^2 - 30t + 9 = 0 \]This is now in the form \( ax^2 + bx + c = 0 \), where \( a = 25 \), \( b = -30 \), and \( c = 9 \).

Step 3: Apply the Quadratic Formula

We use the quadratic formula to solve for \( t \):\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Substitute the values of \( a \), \( b \), and \( c \) into the formula:\[ t = \frac{-(-30) \pm \sqrt{(-30)^2 - 4 \times 25 \times 9}}{2 \times 25} \]

Step 4: Calculate the Discriminant

Calculate the discriminant term inside the square root:\[ b^2 - 4ac = (-30)^2 - 4 \times 25 \times 9 \]\[ = 900 - 900 \]\[ = 0 \]

Step 5: Find the Solutions

Since the discriminant is zero, there is only one solution:\[ t = \frac{30}{50} \]Simplify:\[ t = \frac{3}{5} \]

Step 6: Verify the Solution

Substitute \( t = \frac{3}{5} \) back into the original equation to verify:\[ 5 \times \frac{3}{5} (5 \times \frac{3}{5} - 6) = -9 \]\[ 3(3 - 6) = -9 \]\[ 3(-3) = -9 \]\[ -9 = -9 \]This confirms that the solution \( t = \frac{3}{5} \) is correct.

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations, which are equations of the form \( ax^2 + bx + c = 0 \). These types of equations curve like a parabola on a graph. To solve them using the quadratic formula, you apply this formula:\[t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Here's a breakdown of the components:

• \( a \), \( b \), and \( c \) are coefficients in the quadratic equation.
• \( b^2 - 4ac \) is a key part called the discriminant.
• \( \pm \) signifies that there may be two possible values for \( t \).

By substituting the values of \( a \), \( b \), and \( c \) into this formula, you can find the solutions for \( t \). This method becomes incredibly useful when factoring is not straightforward. Remember that the quadratic formula can be used no matter how complex the numbers might seem. It consistently gives you a reliable method to find solutions.

Discriminant

The discriminant is part of the quadratic formula, specifically located inside the square root. It is expressed as \( b^2 - 4ac \). The discriminant is crucial because it tells us about the nature of the solutions of a quadratic equation.The discriminant, \( b^2 - 4ac \), can lead to different types of solutions based on its value:

• When the discriminant is greater than 0, there are two real and distinct solutions.
• If the discriminant equals 0, there is exactly one real solution; the parabola touches the x-axis only once.
• If the discriminant is less than 0, there are no real solutions, making the solutions complex or imaginary.

In our exercise, the discriminant is 0, which indicates that the equation has just one unique real solution. This is why, after applying the quadratic formula, only one value for \( t \) emerged.

Distributive Property

The distributive property is a key algebraic principle that simplifies expressions and equations by distributing a multiplication over addition or subtraction within parentheses. It is expressed as \( a(b + c) = ab + ac \). This property allows us to expand or simplify expressions as needed.In quadratic equations, the distributive property is essential for expanding terms. As seen in the original equation \( 5t(5t-6) = -9 \), applying the distributive property involves multiplying \( 5t \) with each term inside the parentheses:

• Multiply \( 5t \) by \( 5t \) to get \( 25t^2 \).
• Multiply \( 5t \) by \(-6\) to get \(-30t \).

This gives us the expanded form \( 25t^2 - 30t \). By distributing, you make it easier to manipulate and transform the equation into a standard quadratic form. Understanding the distributive property is fundamental, as it frequently appears in various algebraic processes.