Zeros of a Quadratic Polynomial

Video Explanation

Question

Find the zeros of the quadratic polynomial

\[ p(x) = x^2 + 2\sqrt{2}\,x – 6 \]

and verify the relationship between the zeros and their coefficients.

Solution

Step 1: Factorise the Polynomial

Given:

\[ p(x) = x^2 + 2\sqrt{2}\,x – 6 \]

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

\[ 1 \times (-6) = -6 \]

We split the middle term using \(3\sqrt{2}\) and \(-\sqrt{2}\), since

\[ 3\sqrt{2} – \sqrt{2} = 2\sqrt{2} \quad \text{and} \quad (3\sqrt{2})(-\sqrt{2}) = -6 \]

\[ x^2 + 3\sqrt{2}\,x – \sqrt{2}\,x – 6 \]

Grouping the terms:

\[ (x^2 + 3\sqrt{2}\,x) – (\sqrt{2}\,x + 6) \]

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

\[ (x – \sqrt{2})(x + 3\sqrt{2}) \]

Step 2: Find the Zeros

\[ (x – \sqrt{2})(x + 3\sqrt{2}) = 0 \]

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

\[ x + 3\sqrt{2} = 0 \Rightarrow x = -3\sqrt{2} \]

Hence, the zeros are:

\[ \alpha = \sqrt{2},\quad \beta = -3\sqrt{2} \]

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, \[ a = 1,\; b = 2\sqrt{2},\; c = -6 \]

Sum of the Zeros

\[ \alpha + \beta = \sqrt{2} + (-3\sqrt{2}) = -2\sqrt{2} \]

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

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

Product of the Zeros

\[ \alpha\beta = (\sqrt{2})(-3\sqrt{2}) = -6 \]

\[ \frac{c}{a} = \frac{-6}{1} = -6 \]

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

Conclusion

The zeros of the given quadratic polynomial are:

\[ \sqrt{2} \text{ and } -3\sqrt{2} \]

The relationship between the zeros and the coefficients is verified.

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