Zeros of a Quadratic Polynomial

Video Explanation

Question

Find the zeros of the quadratic polynomial

\[ g(s) = 4s^2 – 4s + 1 \]

and verify the relationship between the zeros and their coefficients.

Solution

Step 1: Find the Zeros of the Polynomial

Given:

\[ g(s) = 4s^2 – 4s + 1 \]

Factorising the polynomial:

\[ 4s^2 – 4s + 1 = (2s – 1)(2s – 1) \]

\[ = (2s – 1)^2 \]

Equating the factor to zero:

\[ 2s – 1 = 0 \]

\[ s = \frac{1}{2} \]

Hence, the zeros of the polynomial are:

\[ \alpha = \frac{1}{2},\quad \beta = \frac{1}{2} \]

Step 2: Verify the Relationship Between Zeros and Coefficients

For a quadratic polynomial \[ as^2 + bs + c, \]

Sum of zeros:

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

Product of zeros:

\[ \alpha\beta = \frac{c}{a} \]

Here, \[ a = 4,\; b = -4,\; c = 1 \]

Sum of the zeros

\[ \alpha + \beta = \frac{1}{2} + \frac{1}{2} = 1 \]

\[ -\frac{b}{a} = -\frac{-4}{4} = 1 \]

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

Product of the zeros

\[ \alpha \beta = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} \]

\[ \frac{c}{a} = \frac{1}{4} \]

\[ \alpha \beta = \frac{c}{a} \quad \checkmark \]

Conclusion

The zeros of the given quadratic polynomial are:

\[ \frac{1}{2} \text{ and } \frac{1}{2} \]

The relationship between the zeros and the coefficients is verified.

\[ \therefore \quad \text{The required result is proved.} \]