Problem 25 - Algebra 1: Concepts and Skills

Question

Solve the equation. Check your solutions. $$ \frac{4}{p}=\frac{p}{16} $$

Step-by-Step Solution

Answer

The solution to the equation \(\frac{4}{p}=\frac{p}{16}\) are \(p = 8\) and \(p = -8\).

Step 1: Cross Multiplication

To eliminate the fraction, cross-multiply the given equation: \(4 \times 16 = p^2\) (p times p). This will give us a simple quadratic equation to solve.

Step 2: Solve Quadratic Equation

Our quadratic equation is \(p^2 = 64\). We solve for \(p\) by taking the square root of both sides. Remember that there are two solutions for \(p\), one positive and one negative: \(p = 8\) and \(p = -8\).

Step 3: Check the Solutions

Substitute \(p = 8\) and \(p = -8\) back into the original equation to ensure both values are valid solutions. For \(p = 8\), we have \(\frac{4}{8} = \frac{8}{16}\) or \(0.5 = 0.5\), which is true. For \(p = -8\), we have \(\frac{4}{-8} = \frac{-8}{16}\) or \(-0.5 = -0.5\), which is also true. Therefore, both 8 and -8 are valid solutions.

Key Concepts

cross multiplication

When dealing with equations involving fractions, cross multiplication is a useful technique. It helps us to eliminate the fractions and transforms the equation into a form that is simpler to solve. Here's how it works.
Let's take the given equation:

• \( \frac{4}{p} = \frac{p}{16} \)

Cross multiplication involves multiplying the numerator of one fraction by the denominator of the other. This operation allows us to "cross" the product across the equal sign:

• Multiply 4 (numerator of first fraction) by 16 (denominator of second fraction).
• Multiply \( p \) (denominator of first fraction) by \( p \) (numerator of the second fraction), resulting in \( p^2 \).

After performing the cross multiplication, we obtain the equation:

• \( 4 \times 16 = p^2 \)
• Which simplifies to \( 64 = p^2 \)

This gives us a neat equation to proceed with, making the next step of solving the quadratic equation straightforward.

quadratic equations

Quadratic equations are equations of the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants. In our exercise, cross multiplication gave us a simple quadratic equation \( p^2 = 64 \). This particular equation lacks the \( b \) and \( c \) terms, making it simpler to solve. To find \( p \), we need to solve for \( p \) by taking the square root of both sides of the equation.
This step involves:

• Recognizing that \( p^2 = 64 \) can be rewritten as \( p = \pm\sqrt{64} \).
• Understanding that the square root of 64 can be both positive 8 and negative 8.

Thus, our solutions become:

• \( p = 8 \)
• \( p = -8 \)

These solutions are vital because they account for all possible scenarios, considering the nature of square roots in quadratics.

checking solutions

Checking solutions is a crucial step in solving equations, as it verifies the validity of the results. After finding potential solutions for \( p \), substituting them back into the original equation confirms their accuracy. This ensures no mistakes were made during the process.
Let's check both solutions:

• For \( p = 8 \), substitute back to get \( \frac{4}{8} = \frac{8}{16} \). Both sides simplify to 0.5, which is true.
• For \( p = -8 \), substitute back to get \( \frac{4}{-8} = \frac{-8}{16} \). Both sides simplify to -0.5, also true.

Since substituting both \( p = 8 \) and \( p = -8 \) into the original equation results in true statements, we can confidently say that both solutions are correct and satisfy the given equation. This step not only confirms the answer but also reinforces the understanding that both positive and negative roots are possible solutions when dealing with quadratic equations.