Zeros of a Quadratic Polynomial

Video Explanation

Question

Find the zeros of the quadratic polynomial

\[ h(s) = 2s^2 – (1 + 2\sqrt{2})s + \sqrt{2} \]

and verify the relationship between the zeros and their coefficients.

Solution

Step 1: Factorise the Polynomial

Given:

\[ h(s) = 2s^2 – (1 + 2\sqrt{2})s + \sqrt{2} \]

Product of coefficient of \(s^2\) and constant term:

\[ 2 \times \sqrt{2} = 2\sqrt{2} \]

We split the middle term using \(-1\) and \(-2\sqrt{2}\), since

\[ -1 – 2\sqrt{2} = -(1 + 2\sqrt{2}) \quad \text{and} \quad (-1)(-2\sqrt{2}) = 2\sqrt{2} \]

\[ 2s^2 – s – 2\sqrt{2}s + \sqrt{2} \]

Grouping the terms:

\[ (2s^2 – s) – (2\sqrt{2}s – \sqrt{2}) \]

\[ s(2s – 1) – \sqrt{2}(2s – 1) \]

\[ (2s – 1)(s – \sqrt{2}) \]

Step 2: Find the Zeros

\[ (2s – 1)(s – \sqrt{2}) = 0 \]

\[ 2s – 1 = 0 \Rightarrow s = \frac{1}{2} \]

\[ s – \sqrt{2} = 0 \Rightarrow s = \sqrt{2} \]

Hence, the zeros are:

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

Step 3: Verify the Relationship Between Zeros and Coefficients

For a quadratic polynomial \(as^2 + bs + c\):

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

Here, \[ a = 2,\; b = -(1 + 2\sqrt{2}),\; c = \sqrt{2} \]

Sum of the Zeros

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

\[ -\frac{b}{a} = -\frac{-(1 + 2\sqrt{2})}{2} = \frac{1 + 2\sqrt{2}}{2} \]

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

Product of the Zeros

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

\[ \frac{c}{a} = \frac{\sqrt{2}}{2} \]

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

Conclusion

The zeros of the given quadratic polynomial are:

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

The relationship between the zeros and the coefficients is verified.

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