Understanding the Algebraic Problem

The question asks us to find the value of the expression \(2p + q\), given two equations involving the variables \(p\) and \(q\). We are given:

1. The linear equation: \(2p + 3q = 18\)
2. The quadratic equation: \(4p^2 + 4pq - 3q^2 - 36 = 0\)

To find the value of \(2p + q\), we first need to determine the values of \(p\) and \(q\) that satisfy both equations simultaneously. This involves solving a system of equations where one is linear and the other is quadratic.

Solving the System of Equations

Let's analyze the second equation: \(4p^2 + 4pq - 3q^2 - 36 = 0\). We can rearrange it as \(4p^2 + 4pq - 3q^2 = 36\).

The left side of this equation, \(4p^2 + 4pq - 3q^2\), is a homogeneous quadratic expression in \(p\) and \(q\). We can try to factor this expression. Let's look for factors in the form \((ap + bq)(cp + dq)\).

We need to find \(a, b, c, d\) such that:

• \(ac = 4\)
• \(bd = -3\)
• \(ad + bc = 4\) (coefficient of \(pq\))

Let's consider possible integer factors for the coefficients. If we try factors for \(4p^2\) and \(-3q^2\), we can test combinations:

• Factors of \(4p^2\): \((p)(4p)\) or \((2p)(2p)\)
• Factors of \(-3q^2\): \((q)(-3q)\) or \((-q)(3q)\)

Let's try the combination \((2p + bq)(2p + dq)\):

\((2p + bq)(2p + dq) = 4p^2 + 2pdq + 2pbq + bdq^2 = 4p^2 + (2d + 2b)pq + bdq^2\)

We need \(bd = -3\) and \(2d + 2b = 4\), which simplifies to \(d + b = 2\).

Possible pairs \((b, d)\) for \(bd = -3\) are \((1, -3), (-1, 3), (3, -1), (-3, 1)\). Let's check if any pair sums to 2:

• \(1 + (-3) = -2\) (No)
• \(-1 + 3 = 2\) (Yes!)
• \(3 + (-1) = 2\) (Yes!)
• \(-3 + 1 = -2\) (No)

So, possible factor pairs for \((b, d)\) are \((-1, 3)\) or \((3, -1)\). This means the factors are \((2p - q)(2p + 3q)\) or \((2p + 3q)(2p - q)\). Let's expand \((2p - q)(2p + 3q)\):

\((2p - q)(2p + 3q) = 2p(2p + 3q) - q(2p + 3q) = 4p^2 + 6pq - 2pq - 3q^2 = 4p^2 + 4pq - 3q^2\)

This matches the quadratic part of the second equation.

So, the second equation \(4p^2 + 4pq - 3q^2 - 36 = 0\) can be written as:

\((2p - q)(2p + 3q) - 36 = 0\)

Which means:

\((2p - q)(2p + 3q) = 36\)

Now, we can use the first given equation, which is \(2p + 3q = 18\). We can substitute this into the factored equation:

\((2p - q) \times (18) = 36\)

Now, we can solve for the expression \((2p - q)\):

\(2p - q = \frac{36}{18}\)

\(2p - q = 2\)

We now have a system of two linear equations:

1. \(2p + 3q = 18\)
2. \(2p - q = 2\)

We can solve this system to find the values of \(p\) and \(q\). Let's subtract the second equation from the first:

This gives us \(4q = 16\). Dividing by 4, we find the value of \(q\):

\(q = \frac{16}{4} = 4\)

Now substitute the value of \(q = 4\) into one of the linear equations, for example, \(2p - q = 2\):

\(2p - 4 = 2\)

Add 4 to both sides:

\(2p = 2 + 4\)

\(2p = 6\)

Divide by 2 to find \(p\):

\(p = \frac{6}{2} = 3\)

We have found \(p = 3\) and \(q = 4\). The question asks for the value of \(2p + q\). Let's substitute the values of \(p\) and \(q\) into this expression:

\(2p + q = 2(3) + 4\)

\(2p + q = 6 + 4\)

\(2p + q = 10\)

Conclusion

By factoring the quadratic equation and using substitution from the linear equation, we determined the values of \(p\) and \(q\) that satisfy both conditions. Substituting these values into the expression \(2p + q\) gives us the final answer.

Revision Table: Key Steps for Solving

Additional Information: Factoring Quadratic Forms

The expression \(4p^2 + 4pq - 3q^2\) is a quadratic form. Factoring such expressions involves finding two linear factors \((ap + bq)\) and \((cp + dq)\) such that their product equals the original expression. This is similar to factoring a standard quadratic expression like \(ax^2 + bx + c\), but with two variables.

The general approach is to look for factors of the coefficients of the squared terms (\(4\) and \(-3\)) and the constant term (\(0\) in this case, before adding the \(-36\)), and arrange them in linear binomials such that the cross-terms add up to the coefficient of the \(pq\) term (\(4\)).

In our case, we factored \(4p^2 + 4pq - 3q^2\) into \((2p - q)(2p + 3q)\). Recognizing this factorization was key to easily solving the system, as it directly utilized the form of the first linear equation.

This method is particularly useful when the constant term in the quadratic equation allows for factorization after isolating the homogeneous quadratic part.