Zeros of a Quadratic Polynomial

Video Explanation

Question

Find the zeros of the quadratic polynomial

\[ f(x) = x^2 – (\sqrt{3} + 1)x + \sqrt{3} \]

and verify the relationship between the zeros and their coefficients.

Solution

Step 1: Factorise the Polynomial

Given:

\[ f(x) = x^2 – (\sqrt{3} + 1)x + \sqrt{3} \]

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

\[ 1 \times \sqrt{3} = \sqrt{3} \]

We split the middle term using \(-\sqrt{3}\) and \(-1\), since

\[ -\sqrt{3} – 1 = -(\sqrt{3} + 1) \quad \text{and} \quad (-\sqrt{3})(-1) = \sqrt{3} \]

\[ x^2 – \sqrt{3}x – x + \sqrt{3} \]

Grouping the terms:

\[ (x^2 – \sqrt{3}x) – (x – \sqrt{3}) \]

\[ x(x – \sqrt{3}) – 1(x – \sqrt{3}) \]

\[ (x – 1)(x – \sqrt{3}) \]

Step 2: Find the Zeros

\[ (x – 1)(x – \sqrt{3}) = 0 \]

\[ x – 1 = 0 \Rightarrow x = 1 \]

\[ x – \sqrt{3} = 0 \Rightarrow x = \sqrt{3} \]

Hence, the zeros are:

\[ \alpha = 1,\quad \beta = \sqrt{3} \]

Step 3: Verify the Relationship Between Zeros and Coefficients

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

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

Here, \[ a = 1,\; b = -(\sqrt{3} + 1),\; c = \sqrt{3} \]

Sum of the Zeros

\[ \alpha + \beta = 1 + \sqrt{3} \]

\[ -\frac{b}{a} = -[-(\sqrt{3} + 1)] = \sqrt{3} + 1 \]

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

Product of the Zeros

\[ \alpha\beta = 1 \times \sqrt{3} = \sqrt{3} \]

\[ \frac{c}{a} = \frac{\sqrt{3}}{1} = \sqrt{3} \]

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

Conclusion

The zeros of the given quadratic polynomial are:

\[ 1 \text{ and } \sqrt{3} \]

The relationship between the zeros and the coefficients is verified.

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