Quadratic Polynomial from Given Zeros

Video Explanation

Question

Find a quadratic polynomial whose sum and product of the zeros are:

\[ \text{Sum} = -\frac{3}{2\sqrt{5}}, \quad \text{Product} = -\frac{1}{2} \]

Also, find the zeros of this polynomial by factorisation.

Solution

Step 1: Use the Standard Form

If the sum of zeros is \(S\) and the product is \(P\), then the quadratic polynomial is:

\[ x^2 – Sx + P \]

Step 2: Form the Required Polynomial

Substitute the given values:

\[ x^2 – \left(-\frac{3}{2\sqrt{5}}\right)x – \frac{1}{2} \]

\[ x^2 + \frac{3}{2\sqrt{5}}x – \frac{1}{2} \]

Multiply the whole polynomial by \(2\sqrt{5}\) to remove fractions:

\[ 2\sqrt{5}x^2 + 3x – \sqrt{5} \]

Hence, the required quadratic polynomial is:

\[ \boxed{2\sqrt{5}x^2 + 3x – \sqrt{5}} \]

Step 3: Factorise the Polynomial

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

\[ 2\sqrt{5} \times (-\sqrt{5}) = -10 \]

Split the middle term using \(5\) and \(-2\):

\[ 2\sqrt{5}x^2 + 5x – 2x – \sqrt{5} \]

Grouping the terms:

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

\[ x(2\sqrt{5}x + 5) – 1(2x + \sqrt{5}) \]

\[ (\sqrt{5}x – 1)(2x + \sqrt{5}) \]

Step 4: Find the Zeros

\[ (\sqrt{5}x – 1)(2x + \sqrt{5}) = 0 \]

\[ \sqrt{5}x – 1 = 0 \Rightarrow x = \frac{1}{\sqrt{5}} \]

\[ 2x + \sqrt{5} = 0 \Rightarrow x = -\frac{\sqrt{5}}{2} \]

Conclusion

The required quadratic polynomial is:

\[ \boxed{2\sqrt{5}x^2 + 3x – \sqrt{5}} \]

The zeros of the polynomial are:

\[ \frac{1}{\sqrt{5}} \quad \text{and} \quad -\frac{\sqrt{5}}{2} \]

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