Condition on Zeros of a Quadratic Polynomial

Video Explanation

Question

If the squared difference of the zeros of the quadratic polynomial

\[ f(x) = x^2 + px + 45 \]

is equal to \(144\), find the value of \(p\).

Solution

Step 1: Write Relations Between Zeros and Coefficients

Let the zeros be \( \alpha \) and \( \beta \).

For a quadratic polynomial \( ax^2 + bx + c \):

\[ \alpha + \beta = -\frac{b}{a}, \quad \alpha\beta = \frac{c}{a} \]

Here,

\[ a = 1,\quad b = p,\quad c = 45 \]

\[ \alpha + \beta = -p,\quad \alpha\beta = 45 \]

Step 2: Use the Formula for Squared Difference of Zeros

\[ (\alpha – \beta)^2 = (\alpha + \beta)^2 – 4\alpha\beta \]

Substitute the values:

\[ (\alpha – \beta)^2 = (-p)^2 – 4(45) \]

\[ (\alpha – \beta)^2 = p^2 – 180 \]

Step 3: Apply the Given Condition

Given:

\[ (\alpha – \beta)^2 = 144 \]

So,

\[ p^2 – 180 = 144 \]

\[ p^2 = 324 \]

\[ p = \pm 18 \]

Conclusion

The required values of \(p\) are:

\[ \boxed{p = 18 \text{ or } p = -18} \]

\[ \therefore \quad p = \pm 18. \]