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Chapter 2: Problem 18

In \(3-20,\) solve each equation and check. $$ \frac{a-1}{4}=\frac{8}{a+3} $$

Short Answer

Expert verified
The solutions are \(a = 5\) and \(a = -7\).

Step by step solution

01

Cross-Multiply to Clear the Fractions

To eliminate the fractions, we can cross-multiply. This means multiplying the numerator of one fraction by the denominator of the other, and vice versa. This will give us the equation:\[(a-1)(a+3) = 4 \times 8\]
02

Expand the Equation

Next, we expand the left-hand side by multiplying out the terms:\[a^2 + 3a - a - 3 = 32\]Simplify the terms:\[a^2 + 2a - 3 = 32\]
03

Bring All Terms to One Side

To set the equation to zero, subtract 32 from both sides:\[a^2 + 2a - 3 - 32 = 0\]This simplifies to:\[a^2 + 2a - 35 = 0\]
04

Solve the Quadratic Equation

Now, solve the quadratic equation using the quadratic formula \(a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Here, \(a = 1\), \(b = 2\), and \(c = -35\):First, calculate the discriminant:\[b^2 - 4ac = 2^2 - 4 \times 1 \times (-35) = 4 + 140 = 144\]Now, apply the quadratic formula:\[a = \frac{-2 \pm \sqrt{144}}{2 \times 1} = \frac{-2 \pm 12}{2}\]This yields two solutions:\[a = \frac{10}{2} = 5\] and \[a = \frac{-14}{2} = -7\]
05

Check the Solutions

Finally, check both possible solutions in the original equation:1. For \(a = 5\): Substitute into the equation \(\frac{a-1}{4} = \frac{8}{a+3}\): \[\frac{5-1}{4} = \frac{8}{5+3}\] or \[\frac{4}{4} = \frac{8}{8}\] which simplifies to \(1 = 1\). This solution is valid.2. For \(a = -7\): Substitute into the equation \(\frac{a-1}{4} = \frac{8}{a+3}\): \[\frac{-7-1}{4} = \frac{8}{-7+3}\] or \[\frac{-8}{4} = \frac{8}{-4}\] which simplifies to \(-2 = -2\). This solution is also valid.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cross-Multiplication
Cross-multiplication is a useful technique often employed to solve equations involving fractions. Here鈥檚 how it works:
  • Imagine you have an equation with two fractions set equal to each other, like \( \frac{a-1}{4} = \frac{8}{a+3} \).
  • To "cross-multiply," multiply the numerator (top) of one fraction by the denominator (bottom) of the other fraction, and do the same with the other numerator and denominator.
  • So, you will have \((a-1)(a+3) = 4 \times 8\).
This method effectively eliminates the fractions, turning the equation into a standard quadratic one, which we can solve more easily with other algebraic methods.
Quadratic Formula
The quadratic formula is an indispensable tool for solving quadratic equations, which take the form \( ax^2 + bx + c = 0 \). The formula is given by:\[a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Let's explore how this formula works:
  • Identify the coefficients \( a \), \( b \), and \( c \) from your quadratic equation. For the equation \( a^2 + 2a - 35 = 0 \), \( a = 1 \), \( b = 2 \), and \( c = -35 \).
  • Plug these values into the quadratic formula to find the solutions of the equation.
  • The solution can result in two different values for \( a \), derived from the \( \pm \) symbol. It provides both possible roots of the equation.
Using the quadratic formula is efficient, especially when factoring is complex or not immediately obvious.
Discriminant Calculation
The discriminant plays a crucial role in understanding the nature of the roots of a quadratic equation. The discriminant \( D \) is the part of the quadratic formula inside the square root:\[b^2 - 4ac\]Here鈥檚 how it helps:
  • If \( D > 0 \), the equation has two distinct real roots.
  • If \( D = 0 \), there is exactly one real root, meaning the parabola touches the x-axis at one point.
  • If \( D < 0 \), there are no real roots; instead, the roots are complex and the parabola does not intersect the x-axis.
In our equation, the discriminant is \( 144 \), which is positive, indicating the presence of two distinct real solutions. Calculating the discriminant helps us anticipate what type of roots 鈥� real or complex 鈥� we should expect before proceeding further.
Solution Verification
Verifying the solutions to a quadratic equation is a pivotal step to ensure correctness. Here鈥檚 what you can do:
  • Once you've found the solutions using the quadratic formula, substitute each value back into the original equation.
  • For instance, check if \( a = 5 \) fits the initial setup \( \frac{a-1}{4} = \frac{8}{a+3} \). It simplifies as \( \frac{4}{4} = \frac{8}{8} \) results in \( 1 = 1 \), confirming it's a valid solution.
  • Do the same with \( a = -7 \). Substitute and simplify to verify this solution. If both sides end equal, it confirms the root's correctness.
This final step not only catches any potential mistakes made during calculations but also reinforces your understanding of the process by confirming the roots are indeed solutions to the original equation.

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Most popular questions from this chapter

In \(21-24,\) the length and width of a rectangle are expressed in terms of a variable. a. Express each perimeter in terms of the variable. b. Express each area in terms of the variable. $$ l=\frac{x}{x+1} \text { and } w=\frac{x}{x+2} $$

Simplify each complex rational expression. In each case, list any values of the variables for which the fractions are not defined. \(\frac{a-\frac{49}{a}}{a-9+\frac{14}{a}}\)

In \(25-30,\) perform the indicated operations and write the result in simplest form. In each case, list any values of the variables for which the fractions are not defined. $$ \frac{3}{5} \times \frac{5}{9} \div \frac{4}{3} $$

In \(3-20\) , perform the indicated additions or subtractions and write the result in simplest form. In each case, list any values of the variables for which the fractions are not defined. $$ 3+\frac{2}{x} $$

In \(3-20\) , perform the indicated additions or subtractions and write the result in simplest form. In each case, list any values of the variables for which the fractions are not defined. $$ \frac{1}{x}+1 $$
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