Problem 74 - Beginning and Intermediate Algebra

Question

The following equations are not quadratic but can be solved by factoring and applying the zero product rule. Solve each equation. $$10 a^{2}(4 a+3)+2(4 a+3)=9 a(4 a+3)$$

Step-by-Step Solution

Answer

The three possible solutions for 'a' are: \(-\frac{3}{4}\), \(\frac{1}{2}\), and \(\frac{2}{5}\).

Step 1: Identify the common factor

Observe the given equation and notice that all terms have a common factor of (4a + 3). $$10a^2(4a+3)+2(4a+3)-9a(4a+3)=0$$

Step 2: Factor out the common factor

Now that we have identified the common factor, let's factor it out of the equation. $$(4a+3)[10a^2+2-9a]=0$$

Step 3: Simplify the equation

Now that we have factored out the common term, we can simplify the equation further. $$(4a+3)(10a^2-9a+2)=0$$

Step 4: Apply the zero product rule

Since the product of the two factors is zero, one or both of them must also be zero. We have the following two possibilities: 1) \(4a + 3 = 0\) 2) \(10a^2 - 9a + 2 = 0\)

Step 5: Solve for 'a' in the first possibility

Let's solve the first possibility for 'a': $$4a + 3 = 0$$ $$4a = -3$$ $$a = -\frac{3}{4}$$

Step 6: Solve for 'a' in the second possibility

Since the second possibility is a quadratic equation, we will solve for 'a' using the quadratic formula: $$10a^2 - 9a + 2 = 0$$ The quadratic formula is: $$a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ For our equation, a = 10, b = -9, and c = 2.

Step 7: Calculate 'a' using the quadratic formula

Plugging the values into the quadratic formula, we get: $$a = \frac{9 \pm \sqrt{(-9)^2 - 4(10)(2)}}{2(10)}$$ $$a = \frac{9 \pm \sqrt{81 - 80}}{20}$$ $$a = \frac{9 \pm \sqrt{1}}{20}$$ Now, let's find the two possible values for 'a':

Step 8: Calculate both possible values for 'a'

First, consider the positive square root: $$a = \frac{9 + 1}{20}$$ $$a = \frac{10}{20}$$ $$a = \frac{1}{2}$$ Next, consider the negative square root: $$a = \frac{9 - 1}{20}$$ $$a = \frac{8}{20}$$ $$a = \frac{2}{5}$$

Step 9: Final Answer

So the three possible solutions for 'a' are: 1) \(-\frac{3}{4}\) 2) \(\frac{1}{2}\) 3) \(\frac{2}{5}\)

Key Concepts

Zero Product Rule

The Zero Product Rule is a handy tool in algebra. It states that if the product of two or more factors is zero, then at least one of the factors must be zero. In simpler terms, if you have something like \( x \times y = 0 \), either \( x = 0 \) or \( y = 0 \), or both. This rule comes in very useful when solving equations after factoring. For instance, in our problem, after factoring the expression, we had:

• \((4a+3)(10a^2-9a+2) = 0\)

Breaking it down using the Zero Product Rule, you find separate, simpler equations:

• \(4a + 3 = 0\)
• \(10a^2 - 9a + 2 = 0\)

Solving each one individually gives potential solutions for \( a \). Remember: Anytime you see a product equal to zero, think Zero Product Rule!

Quadratic Equations

Quadratic equations are equations of the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants. These equations often come up in algebra, and they can describe many real-world phenomena, like the path of a projectile or the area of a plot of land. In our problem, we found a quadratic equation within the broader expression:

• \(10a^2 - 9a + 2 = 0\)

To solve this, we used the quadratic formula:\[a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]With the values of \( a = 10 \), \( b = -9 \), and \( c = 2 \), we calculated the possible values of \( a \). The quadratic formula is super powerful and will work every time, as long as the discriminant \( (b^2 - 4ac) \) is non-negative.

Common Factor Identification

The first step in solving many algebraic equations, like the one we're dealing with here, is to identify any common factors present in the terms. This is crucial for simplifying and ultimately solving the equation. For instance, in our given equation:

• \(10a^2(4a+3) + 2(4a+3) - 9a(4a+3) = 0\)

Notice the repetition? \((4a+3)\) appears in every term! By recognizing this common factor, you can factor it out to make the equation simpler:

• \((4a+3)[10a^2+2-9a] = 0\)

This makes it easier to apply rules and solve. Identifying common factors saves time and simplifies the solving process, turning a complex-looking equation into something much more manageable.