Zeros of a Quadratic Polynomial

Video Explanation

Question

Find the zeros of the quadratic polynomial

\[ f(v) = v^2 + 4\sqrt{3}v – 15 \]

and verify the relationship between the zeros and their coefficients.

Solution

Step 1: Factorise the Polynomial

Given:

\[ f(v) = v^2 + 4\sqrt{3}v – 15 \]

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

\[ 1 \times (-15) = -15 \]

We split the middle term using \(5\sqrt{3}\) and \(-\sqrt{3}\), since

\[ 5\sqrt{3} – \sqrt{3} = 4\sqrt{3} \quad \text{and} \quad (5\sqrt{3})(-\sqrt{3}) = -15 \]

\[ v^2 + 5\sqrt{3}v – \sqrt{3}v – 15 \]

Grouping the terms:

\[ (v^2 + 5\sqrt{3}v) – (\sqrt{3}v + 15) \]

\[ v(v + 5\sqrt{3}) – \sqrt{3}(v + 5\sqrt{3}) \]

\[ (v – \sqrt{3})(v + 5\sqrt{3}) \]

Step 2: Find the Zeros

\[ (v – \sqrt{3})(v + 5\sqrt{3}) = 0 \]

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

\[ v + 5\sqrt{3} = 0 \Rightarrow v = -5\sqrt{3} \]

Hence, the zeros are:

\[ \alpha = \sqrt{3},\quad \beta = -5\sqrt{3} \]

Step 3: Verify the Relationship Between Zeros and Coefficients

For a quadratic polynomial \(av^2 + bv + c\):

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

Here, \[ a = 1,\; b = 4\sqrt{3},\; c = -15 \]

Sum of the Zeros

\[ \alpha + \beta = \sqrt{3} + (-5\sqrt{3}) = -4\sqrt{3} \]

\[ -\frac{b}{a} = -\frac{4\sqrt{3}}{1} = -4\sqrt{3} \]

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

Product of the Zeros

\[ \alpha\beta = (\sqrt{3})(-5\sqrt{3}) = -15 \]

\[ \frac{c}{a} = \frac{-15}{1} = -15 \]

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

Conclusion

The zeros of the given quadratic polynomial are:

\[ \sqrt{3} \text{ and } -5\sqrt{3} \]

The relationship between the zeros and the coefficients is verified.

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