Problem 52 - Algebra for College Students

Question

Solve each quadratic equation using the method that seems most appropriate. $$t(t-26)=-160$$

Step-by-Step Solution

Answer

The solutions are \( t = 16 \) and \( t = 10 \).

Step 1: Set the Equation to Zero

First, we need to rewrite the equation in the standard quadratic form. The given equation is \(t(t-26) = -160\). To set it to zero, add 160 to both sides. This gives us:\[ t^2 - 26t + 160 = 0 \]

Step 2: Apply the Quadratic Formula

The standard quadratic equation is \( at^2 + bt + c = 0 \), with \( a = 1 \), \( b = -26 \), and \( c = 160 \). The quadratic formula is \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Plug the values into the formula:\( b^2 - 4ac = (-26)^2 - 4(1)(160) \)Calculate \( b^2 - 4ac \):\( 676 - 640 = 36 \)Now plug \( b = -26 \) and \( \sqrt{36} = 6 \) into the quadratic formula:\[ t = \frac{-(-26) \pm 6}{2 \times 1} \]

Step 3: Calculate the Possible Values of t

Complete the calculations from the quadratic formula:\[ t = \frac{26 \pm 6}{2} \]This gives two solutions:1. \( t = \frac{26 + 6}{2} = \frac{32}{2} = 16 \)2. \( t = \frac{26 - 6}{2} = \frac{20}{2} = 10 \)

Step 4: Verify the Solutions

Insert each solution back into the original equation to verify:1. For \( t = 16 \): - \( 16(16-26) = 16(-10) = -160 \), which is correct.2. For \( t = 10 \): - \( 10(10-26) = 10(-16) = -160 \), which is also correct.Both solutions are verified.

Key Concepts

Quadratic Formula

When faced with a quadratic equation, one of the most powerful tools at your disposal is the quadratic formula. This formula provides a direct method to find the solutions (or roots) of any quadratic equation, which is in the standard form:\[ ax^2 + bx + c = 0 \]Here, \( a \), \( b \), and \( c \) are constants, with \( a eq 0 \). The quadratic formula is expressed as:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]What makes the quadratic formula particularly useful is its completeness. It can find solutions even when other methods fall short, such as when the equation doesn’t factor neatly or graph methods are impractical. Remember, the term inside the square root, \( b^2 - 4ac \), is called the discriminant. This part tells you a bit more about the nature of the solutions:

• If the discriminant is positive, you'll have two distinct real solutions.
• If it's zero, there's only one real solution (the parabola touches the x-axis).
• A negative discriminant means there are no real solutions — the roots are complex.

By substituting the values of \( b \), \( a \), and \( c \) into the formula, you step directly towards solving the equation. This method is thorough and reliable, making it a go-to technique for mathematicians.

Solving Quadratic Equations

Solving quadratic equations is fundamental in algebra, and it often involves finding the values of the variable that satisfy the equation. Consider the equation we have: \[ t(t-26) = -160 \]The first step is to rewrite this equation in the standard quadratic form by expanding and rearranging terms. Doing so gives us:\[ t^2 - 26t + 160 = 0 \]Now, this is ready to be tackled by the quadratic formula. Identify \( a = 1 \), \( b = -26 \), and \( c = 160 \). Plug these into the quadratic formula:\[ t = \frac{-(-26) \pm \sqrt{(-26)^2 - 4 \cdot 1 \cdot 160}}{2 \cdot 1} \]Calculate the discriminant, which is \( 36 \) (because \( 676 - 640 = 36 \)), confirming that there are two real solutions since it's positive. Applying these into the formula results in:

• First solution: \( t = \frac{26 + 6}{2} = 16 \)
• Second solution: \( t = \frac{26 - 6}{2} = 10 \)

This method not only helps find quick solutions but also reinforces understanding of quadratic behavior. Practice by modifying coefficients and observing changes in solutions to grasp different solution scenarios.

Verification of Solutions

Verifying solutions to quadratic equations ensures that the solutions found are accurate and satisfy the original equation. After finding the potential solutions, substitution back into the original equation provides confirmation.Take the solutions \( t = 16 \) and \( t = 10 \) obtained from our earlier calculations. Insert them individually:

• For \( t = 16 \): Substitute back into the equation to check.\[ 16(16 - 26) = 16(-10) = -160 \]This is true, confirming \( t = 16 \) is a valid solution.
• For \( t = 10 \): Verify similarly.\[ 10(10 - 26) = 10(-16) = -160 \]This holds true as well, verifying \( t = 10 \) as another valid solution.

This verification process can be done quickly by testing each root. It affirms that both solutions truly solve the original equation. This step is crucial in mathematical problem-solving, ensuring that no mistakes have been made during calculation. It reinforces trust in the results obtained and solidifies methodological understanding.