\frac{n}{2} (4n + 10) = 150 \implies n(4n + 10) = 300 \implies 4n^2 + 10n - 300 = 0
Solving the Quadratic Equation: \frac{n}{2}(4n + 10) = 150 – Step-by-Step Guide
If you've ever encountered an equation like \(\frac{n}{2}(4n + 10) = 150\), you know how powerful algebra can be when solving for unknown variables. This article breaks down how to solve this quadratic equation step by step, showing how to transform, simplify, and apply the quadratic formula to find accurate values of \(n\).
Understanding the Equation
We begin with:
\[\frac{n}{2}(4n + 10) = 150\]
This equation suggests a proportional relationship multiplied by a linear expression, then set equal to a constant. Solving this will help us uncover the value(s) of \(n\) that satisfy the equation.
Step 1: Eliminate the fraction
To simplify, multiply both sides of the equation by 2:
\[2 \cdot \frac{n}{2}(4n + 10) = 2 \cdot 150\]
\[n(4n + 10) = 300\]
Now we expand the left-hand side.
Step 2: Expand the quadratic expression
Distribute \(n\) across the parentheses:
\[n \cdot 4n + n \cdot 10 = 4n^2 + 10n\]
So the equation becomes:
\[4n^2 + 10n = 300\]
Step 3: Bring all terms to one side
To form a standard quadratic equation, subtract 300 from both sides:
\[4n^2 + 10n - 300 = 0\]
This is now a standard quadratic equation in the form \(an^2 + bn + c = 0\), where:
- \(a = 4\)- \(b = 10\)- \(c = -300\)
Step 4: Simplify the equation (optional but helpful)
Before applying the quadratic formula, we can simplify by dividing the entire equation by the greatest common divisor of the coefficients. Here, \( \gcd(4, 10, 300) = 2 \):
\[\frac{4n^2 + 10n - 300}{2} = 0 \implies 2n^2 + 5n - 150 = 0\]
This simplified form is easier to work with, though solving either \(4n^2 + 10n - 300 = 0\) or \(2n^2 + 5n - 150 = 0\) yields the same solutions.
Step 5: Apply the quadratic formula
For the equation \(an^2 + bn + c = 0\), the solutions are:
\[n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
Using \(2n^2 + 5n - 150 = 0\), identify:
- \(a = 2\)- \(b = 5\)- \(c = -150\)
Compute the discriminant:
\[\Delta = b^2 - 4ac = 5^2 - 4(2)(-150) = 25 + 1200 = 1225\]
Take the square root:
\[\sqrt{1225} = 35\]
Now substitute into the formula:
\[n = \frac{-5 \pm 35}{2 \cdot 2} = \frac{-5 \pm 35}{4}\]
This gives two solutions:
-
\(n = \frac{-5 + 35}{4} = \frac{30}{4} = 7.5\)
-
\(n = \frac{-5 - 35}{4} = \frac{-40}{4} = -10\)
Step 6: Interpret the solutions
In most practical contexts, only positive values make sense for variables like \(n\), especially if it represents a physical quantity or count. Therefore, we discard the negative solution \(n = -10\).
Thus, the valid solution is:
\[n = 7.5\]
Summary
Solving \(\frac{n}{2}(4n + 10) = 150\) involves:
- Eliminating fractions2. Expanding the expression3. Rearranging into standard quadratic form4. Simplifying (optional but recommended)5. Applying the quadratic formula6. Interpreting real-world feasibility
The equation reduces to:
\[4n^2 + 10n - 300 = 0 \quad \ ext{or} \quad 2n^2 + 5n - 150 = 0\]
and yields \(n = 7.5\) as the meaningful solution.
Tips for Tackling Similar Problems
- Always simplify equations before solving- Watch for factoring opportunities, though quadratics often require the quadratic formula- Check solutions by plugging them back into the original equation- Consider context to validate whether multiple or positive roots make sense
Understanding how to solve quadratic equations like \(\frac{n}{2}(4n + 10) = 150\) equips you with clear, logical steps applicable across scientific, financial, and engineering problems where proportional relationships model reality. Try practicing with different values to strengthen your algebra skills!
Keywords: solve quadratic equation, 4n² + 10n - 300 = 0, quadratic formula solution, algebra step-by-step, n = 7.5, solve n(4n + 10) = 150
