The equations 3x2−5x+p=0 and 2x2−2x+q=0 have one common root. The sum of the other roots of these two equations is
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Official Correct Answer: C. First, the equations are:
1. \(3x^2 - 5x + p = 0\)
2. \(2x^2 - 2x + q = 0\)
They share one common root. Let's call that common root \( \alpha \). So, \( \alpha \) satisfies both equations.
for a quadratic equation \( ax^2 + bx + c = 0 \), the sum of the roots is \( -b/a \) and the product is \( c/a \). Maybe I can use that here.
Let the roots of the first equation as \( \alpha \) and \( \beta \). So, the sum of the roots \( \alpha + \beta = \frac{5}{3} \) and the product \( \alpha \beta = \frac{p}{3} \).
Similarly, for the second equation, let the roots be \( \alpha \) and \( \gamma \). So, the sum \( \alpha + \gamma = \frac{2}{2} = 1 \) and the product \( \alpha \gamma = \frac{q}{2} \).
We find the sum of the other roots, which would be \( \beta + \gamma \).
Wait, how can I find \( \beta + \gamma \)? .
From the first equation, \( \beta = \frac{5}{3} - \alpha \).
From the second equation, \( \gamma = 1 - \alpha \).
So, adding them together: \( \beta + \gamma = \left( \frac{5}{3} - \alpha \right) + \left( 1 - \alpha \right) = \frac{5}{3} + 1 - 2\alpha = \frac{8}{3} - 2\alpha \).
Hmm, but I don't know the value of \( \alpha \). Maybe I can find \( \alpha \) by using the fact that it satisfies both equations.
Let me set up the two equations:
1. \(3\alpha^2 - 5\alpha + p = 0\) => \(3\alpha^2 = 5\alpha - p\)
2. \(2\alpha^2 - 2\alpha + q = 0\) => \(2\alpha^2 = 2\alpha - q\)
I can solve for \( \alpha^2 \) in both and set them equal.
From equation 1: \( \alpha^2 = \frac{5\alpha - p}{3} \)
From equation 2: \( \alpha^2 = \frac{2\alpha - q}{2} \)
So, \( \frac{5\alpha - p}{3} = \frac{2\alpha - q}{2} \)
Cross-multiplying: \( 2(5\alpha - p) = 3(2\alpha - q) \)
Expanding both sides: \(10\alpha - 2p = 6\alpha - 3q\)
Bring like terms together: \(10\alpha - 6\alpha = 2p - 3q\)
Simplify: \(4\alpha = 2p - 3q\) => \( \alpha = \frac{2p - 3q}{4} \)
Now, plug this back into the expression for \( \beta + \gamma \):
\( \beta + \gamma = \frac{8}{3} - 2\alpha = \frac{8}{3} - 2 \left( \frac{2p - 3q}{4} \right) \)
Simplify the second term: \( 2 \times \frac{2p - 3q}{4} = \frac{2p - 3q}{2} \)
So, \( \beta + \gamma = \frac{8}{3} - \frac{2p - 3q}{2} \)
To combine these, let's get a common denominator, which is 6:
Convert \( \frac{8}{3} \) to \( \frac{16}{6} \) and \( \frac{2p - 3q}{2} \) to \( \frac{6p - 9q}{6} \)
So, \( \beta + \gamma = \frac{16}{6} - \frac{6p - 9q}{6} = \frac{16 - 6p + 9q}{6} \)
Simplify numerator: \(16 - 6p + 9q\)
Wait, looking at the options, none of them have a denominator of 6. if I can factor this differently or perhaps I made a mistake in the calculation.
From \( \beta + \gamma = \frac{8}{3} - 2\alpha \), and \( \alpha = \frac{2p - 3q}{4} \), so:
\( 2\alpha = \frac{2p - 3q}{2} \)
Thus, \( \beta + \gamma = \frac{8}{3} - \frac{2p - 3q}{2} \)
To combine these, let's express both terms with denominator 6:
\( \frac{8}{3} = \frac{16}{6} \) and \( \frac{2p - 3q}{2} = \frac{6p - 9q}{6} \)
So, \( \beta + \gamma = \frac{16}{6} - \frac{6p - 9q}{6} = \frac{16 - 6p + 9q}{6} \)
Hmm, that's \( \frac{16 - 6p + 9q}{6} \). Factoring numerator:
Factor numerator: 16 -6p +9q = 16 + (-6p +9q) = 16 + 3(-2p + 3q)
Not sure if that helps. the options:
Option C is \( \frac{8}{3} - p + \frac{3}{2}q \)
Compute \( \frac{8}{3} - p + \frac{3}{2}q \) and see if it's equal to \( \frac{16 -6p +9q}{6} \).
Compute \( \frac{8}{3} - p + \frac{3}{2}q \):
Convert all terms to sixths:
\( \frac{8}{3} = \frac{16}{6} \), \( p = \frac{6p}{6} \), \( \frac{3}{2}q = \frac