Question

If \(\alpha\) and \(\beta\) are two zeroes of the quadratic polynomial \(p(x) = x^2 - 11x + 30\), then \(\frac{1}{\alpha} + \frac{1}{\beta}\) is equal to :

• A. \(\frac{30}{11}\)
• B. \(\frac{11}{30}\)
• C. \(-\frac{11}{30}\)
• D. \(-\frac{30}{11}\)

Hint:

Whenever you are asked to evaluate symmetric expressions of zeroes like \(\frac{1}{\alpha} + \frac{1}{\beta}\) or \(\alpha^2 + \beta^2\), always try to express them in terms of \(\alpha + \beta\) and \(\alpha\beta\).
This avoids the need to find the individual roots \(\alpha\) and \(\beta\) of the quadratic equation, which can sometimes be complex or irrational, thereby saving significant calculation time.

Answer 1

The Correct Option is B

Step 1: Understand what is being asked.
We need the value of $\frac{1}{\alpha}+\frac{1}{\beta}$ where $\alpha$ and $\beta$ are the zeroes of $p(x)=x^2-11x+30$. Instead of using the sum and product formulas, let us find the actual zeroes by factoring the polynomial.

Step 2: Factor the quadratic polynomial.
We need two numbers whose sum is 11 and whose product is 30, since the polynomial is $x^2 - 11x + 30$.
The numbers 5 and 6 satisfy this, since $5+6=11$ and $5 \times 6=30$.
So we can write:
\[ x^2 - 11x + 30 = (x-5)(x-6) \]
Step 3: Find the zeroes.
Setting each factor to zero gives the roots:
\[ x - 5 = 0 \implies x = 5 \]
\[ x - 6 = 0 \implies x = 6 \]
So we can take $\alpha = 5$ and $\beta = 6$ (the labelling does not matter here).

Step 4: Substitute the actual root values into the required expression.
\[ \frac{1}{\alpha} + \frac{1}{\beta} = \frac{1}{5} + \frac{1}{6} \]
Take the LCM of 5 and 6, which is 30:
\[ \frac{1}{5} + \frac{1}{6} = \frac{6}{30} + \frac{5}{30} = \frac{11}{30} \]
Final Answer:
The value of $\frac{1}{\alpha}+\frac{1}{\beta}$ is $\frac{11}{30}$, matching option (B).
\[ \boxed{\frac{11}{30}} \]