Solving Algebra Problem: Finding 3m + n

This problem asks us to find the value of the expression \(3m + n\), given two equations involving positive real numbers \(m\) and \(n\). The given equations are:

1. \(9m^2 + n^2 = 40\)
2. \(mn = 4\)

We need to find the value of \(3m + n\). Let's consider squaring the expression we want to find, \( (3m + n)^2 \). Expanding this using the formula \( (a+b)^2 = a^2 + b^2 + 2ab \), we get:

\( (3m + n)^2 = (3m)^2 + (n)^2 + 2(3m)(n) \)

Simplifying the terms:

\( (3m + n)^2 = 9m^2 + n^2 + 6mn \)

Now, notice that the expanded form \( 9m^2 + n^2 + 6mn \) contains the terms from our given equations: \( 9m^2 + n^2 \) and \( mn \). We can substitute the values from the given equations into this expression.

From equation (1), we know \( 9m^2 + n^2 = 40 \).

From equation (2), we know \( mn = 4 \).

Substitute these values into the expanded equation for \( (3m + n)^2 \):

\( (3m + n)^2 = (9m^2 + n^2) + 6(mn) \)

\( (3m + n)^2 = 40 + 6(4) \)

Perform the multiplication:

\( (3m + n)^2 = 40 + 24 \)

Add the numbers:

\( (3m + n)^2 = 64 \)

Now we have the value of \( (3m + n)^2 \). To find the value of \( 3m + n \), we need to take the square root of both sides:

\( 3m + n = \sqrt{64} \)

The square root of 64 is either +8 or -8.

\( 3m + n = \pm 8 \)

However, the problem states that \(m\) and \(n\) are positive real numbers.

• If \(m > 0\) and \(n > 0\), then \(3m\) will also be positive (\(3m > 0\)).
• The sum of two positive numbers (\(3m\) and \(n\)) must be positive.

Therefore, the value of \(3m + n\) must be the positive square root.

\( 3m + n = 8 \)

This matches one of the given options.

Step-by-Step Calculation of 3m + n

1. Start with the expression to find: \(3m + n\).
2. Square the expression: \( (3m + n)^2 \).
3. Expand the squared expression: \( (3m + n)^2 = 9m^2 + n^2 + 6mn \).
4. Substitute the given values \(9m^2 + n^2 = 40\) and \(mn = 4\).
5. \( (3m + n)^2 = 40 + 6(4) \).
6. Calculate \(6 \times 4 = 24\).
7. \( (3m + n)^2 = 40 + 24 \).
8. Calculate \(40 + 24 = 64\).
9. \( (3m + n)^2 = 64 \).
10. Take the square root: \( 3m + n = \sqrt{64} \).
11. Consider both positive and negative roots: \( \pm 8 \).
12. Use the condition that \(m\) and \(n\) are positive real numbers, implying \(3m + n\) must be positive.
13. The value is \( 3m + n = 8 \).

Revision Table: Key Concepts

Additional Information: Solving Simultaneous Equations

While this problem was solved using an algebraic identity, sometimes similar problems might require solving the system of equations to find \(m\) and \(n\) first.

We have the equations:

1. \(9m^2 + n^2 = 40\)
2. \(mn = 4\)

From equation (2), we can express \(n\) in terms of \(m\): \( n = \frac{4}{m} \).

Since \(m\) is a positive real number, \(m \neq 0\).

Substitute this expression for \(n\) into equation (1):

\( 9m^2 + \left(\frac{4}{m}\right)^2 = 40 \)

\( 9m^2 + \frac{16}{m^2} = 40 \)

Multiply the entire equation by \(m^2\) to eliminate the denominator:

\( m^2(9m^2) + m^2\left(\frac{16}{m^2}\right) = 40m^2 \)

\( 9m^4 + 16 = 40m^2 \)

Rearrange the terms to form a quadratic equation in terms of \(m^2\):

\( 9m^4 - 40m^2 + 16 = 0 \)

Let \(x = m^2\). Since \(m\) is a real number, \(m^2\) must be non-negative.

\( 9x^2 - 40x + 16 = 0 \)

We can solve this quadratic equation for \(x\) using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \(a=9\), \(b=-40\), \(c=16\).

\( x = \frac{-(-40) \pm \sqrt{(-40)^2 - 4(9)(16)}}{2(9)} \)

\( x = \frac{40 \pm \sqrt{1600 - 576}}{18} \)

\( x = \frac{40 \pm \sqrt{1024}}{18} \)

\( x = \frac{40 \pm 32}{18} \)

Two possible values for \(x\):

\( x_1 = \frac{40 + 32}{18} = \frac{72}{18} = 4 \)

\( x_2 = \frac{40 - 32}{18} = \frac{8}{18} = \frac{4}{9} \)

Since \(x = m^2\), we have:

Case 1: \( m^2 = 4 \). Since \(m\) is positive, \(m = \sqrt{4} = 2\). If \(m = 2\), then from \(mn = 4\), \(2n = 4\), so \(n = 2\). Check if \(m=2, n=2\) satisfy \(9m^2 + n^2 = 40\): \(9(2^2) + 2^2 = 9(4) + 4 = 36 + 4 = 40\). This solution works. In this case, \(3m + n = 3(2) + 2 = 6 + 2 = 8\).

Case 2: \( m^2 = \frac{4}{9} \). Since \(m\) is positive, \(m = \sqrt{\frac{4}{9}} = \frac{2}{3}\). If \(m = \frac{2}{3}\), then from \(mn = 4\), \(\left(\frac{2}{3}\right)n = 4\). Multiply by \(\frac{3}{2}\): \( n = 4 \times \frac{3}{2} = 6 \). Check if \(m=\frac{2}{3}, n=6\) satisfy \(9m^2 + n^2 = 40\): \(9\left(\left(\frac{2}{3}\right)^2\right) + 6^2 = 9\left(\frac{4}{9}\right) + 36 = 4 + 36 = 40\). This solution also works. In this case, \(3m + n = 3\left(\frac{2}{3}\right) + 6 = 2 + 6 = 8\).

In both valid cases for positive \(m\) and \(n\), the value of \(3m + n\) is 8. This confirms the result obtained using the algebraic identity method. The identity method was faster for this specific problem because the expression to be found \( (3m+n) \) squared conveniently related to the given terms.