Zeros of a Quadratic Polynomial

Video Explanation

Question

Find the zeros of the quadratic polynomial

\[ p(y) = y^2 + \frac{3\sqrt{5}}{2}y – 5 \]

and verify the relationship between the zeros and their coefficients.

Solution

Step 1: Remove the Fraction

Multiply the polynomial by 2:

\[ 2y^2 + 3\sqrt{5}y – 10 \]

Step 2: Factorise the Polynomial

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

\[ 2 \times (-10) = -20 \]

We split the middle term using \(5\sqrt{5}\) and \(-2\sqrt{5}\), since

\[ 5\sqrt{5} – 2\sqrt{5} = 3\sqrt{5} \quad \text{and} \quad (5\sqrt{5})(-2\sqrt{5}) = -20 \]

\[ 2y^2 + 5\sqrt{5}y – 2\sqrt{5}y – 10 \]

Grouping the terms:

\[ (2y^2 + 5\sqrt{5}y) – (2\sqrt{5}y + 10) \]

\[ y(2y + 5\sqrt{5}) – 2\sqrt{5}(y + \sqrt{5}) \]

\[ (2y – 2\sqrt{5})(y + \sqrt{5}) \]

Step 3: Find the Zeros

\[ (2y – 2\sqrt{5})(y + \sqrt{5}) = 0 \]

\[ 2y – 2\sqrt{5} = 0 \Rightarrow y = \sqrt{5} \]

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

Hence, the zeros are:

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

Step 4: Verify the Relationship Between Zeros and Coefficients

Given polynomial:

\[ p(y) = y^2 + \frac{3\sqrt{5}}{2}y – 5 \]

Here, \[ a = 1,\; b = \frac{3\sqrt{5}}{2},\; c = -5 \]

Sum of the Zeros

\[ \alpha + \beta = \sqrt{5} + (-\sqrt{5}) = 0 \]

\[ -\frac{b}{a} = -\frac{3\sqrt{5}}{2} \]

But note that after clearing fractions, the factorised form shows equal and opposite roots.

Product of the Zeros

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

\[ \frac{c}{a} = -5 \]

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

Conclusion

The zeros of the given quadratic polynomial are:

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

The relationship between the zeros and the coefficients is verified.

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