Finding Constants Using Zeroes of a Quadratic Polynomial

Video Explanation

Question

If the zeroes of the quadratic polynomial

\[ f(x) = x^2 + (a + 1)x + b \]

are \(2\) and \(-3\), find the values of \(a\) and \(b\).

Options:

(a) \(a = -7,\; b = -1\)
(b) \(a = 5,\; b = -1\)
(c) \(a = 2,\; b = -6\)
(d) \(a = 0,\; b = -6\)

Solution

Step 1: Use Relations Between Zeroes and Coefficients

For a quadratic polynomial \[ x^2 + px + q, \]

Sum of zeroes \(= -p\)
Product of zeroes \(= q\)

Step 2: Apply to the Given Polynomial

Given polynomial:

\[ x^2 + (a + 1)x + b \]

Sum of zeroes:

\[ 2 + (-3) = -1 \]

So,

\[ -(a + 1) = -1 \Rightarrow a + 1 = 1 \Rightarrow a = 0 \]

Step 3: Find the Value of \(b\)

Product of zeroes:

\[ 2 \times (-3) = -6 \]

So,

\[ b = -6 \]

Conclusion

The values of \(a\) and \(b\) are:

\[ \boxed{a = 0,\; b = -6} \]

Hence, the correct option is (d).