If `3p^(2)=5p+2` and `3q^(2)=5q+2` then the equation whose roots `3p-2p` and `3q-2p` is

Options

A. `x^(2)-5x+100=0`
B. `3x^(2)-5x-100=0`
C. `3x^(2)+5x+100=0`
D. `5x^(2)-x+7=0`

Explanation

To solve the problem, we need to find the equation whose roots are \(3p - 2q\) and \(3q - 2p\), given the equations \(3p^2 = 5p + 2\) and \(3q^2 = 5q + 2\).

### Step 1: Solve for \(p\) and \(q\)

First, we need to rearrange the equations for \(p\) and \(q\).

1. For \(p\):
\[ 3p^2 - 5p - 2 = 0 \]
We can use the quadratic formula \(p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[ a = 3, \quad b = -5, \quad c = -2 \]
\[ p = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 3 \cdot (-2)}}{2 \cdot 3} \]
\[ = \frac{5 \pm \sqrt{25 + 24}}{6} = \frac{5 \pm \sqrt{49}}{6} = \frac{5 \pm 7}{6} \]
This gives us:
\[ p_1 = \frac{12}{6} = 2, \quad p_2 = \frac{-2}{6} = -\frac{1}{3} \]

2. For \(q\):
\[ 3q^2 - 5q - 2 = 0 \]
Using the quadratic formula again:
\[ q = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 3 \cdot (-2)}}{2 \cdot 3} \]
This gives us the same roots as for \(p\):
\[ q_1 = 2, \quad q_2 = -\frac{1}{3} \]

### Step 2: Calculate the roots \(3p - 2q\) and \(3q - 2p\)

Now, we need to find the roots \(3p - 2q\) and \(3q - 2p\).

1. For \(p = 2\) and \(q = 2\):
\[ 3p - 2q = 3(2) - 2(2) = 6 - 4 = 2 \]
\[ 3q - 2p = 3(2) - 2(2) = 6 - 4 = 2 \]

2. For \(p = 2\) and \(q = -\frac{1}{3}\):
\[ 3p - 2q = 3(2) - 2\left(-\frac{1}{3}\right) = 6 + \frac{2}{3} = \frac{18}{3} + \frac{2}{3} = \frac{20}{3} \]
\[ 3q - 2p = 3\left(-\frac{1}{3}\right) - 2(2) = -1 - 4 = -5 \]

3. For \(p = -\frac{1}{3}\) and \(q = 2\):
\[ 3p - 2q = 3\left(-\frac{1}{3}\right) - 2(2) = -1 - 4 = -5 \]
\[ 3q - 2p = 3(2) - 2\left(-\frac{1}{3}\right) = 6 + \frac{2}{3} = \frac{20}{3} \]

### Step 3: Form the equation with roots

Now we have the roots:
1. \(2\) and \(2\)
2. \(\frac{20}{3}\) and \(-5\)

Using the roots \(\frac{20}{3}\) and \(-5\), we can find the equation.

The sum of the roots:
\[ \alpha + \beta = \frac{20}{3} + (-5) = \frac{20}{3} - \frac{15}{3} = \frac{5}{3} \]

The product of the roots:
\[ \alpha \beta = \frac{20}{3} \cdot (-5) = -\frac{100}{3} \]

Using the standard form of a quadratic equation \(x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0\):
\[ x^2 - \frac{5}{3}x - \frac{100}{3} = 0 \]

Multiplying through by 3 to eliminate the fractions:
\[ 3x^2 - 5x - 100 = 0 \]

### Final Answer

The equation whose roots are \(3p - 2q\) and \(3q - 2p\) is:
\[ \boxed{3x^2 - 5x - 100 = 0} \]