Zeros of a Quadratic Polynomial

Video Explanation

Question

Find the zeros of the quadratic polynomial

\[ q(y) = 7y^2 – \frac{11}{3}y – \frac{2}{3} \]

and verify the relationship between the zeros and their coefficients.

Solution

Step 1: Remove Fractions

Multiply the polynomial by 3:

\[ 3q(y) = 21y^2 – 11y – 2 \]

Step 2: Factorise the Polynomial

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

\[ 21 \times (-2) = -42 \]

We split the middle term using \(-14\) and \(+3\), since

\[ -14 + 3 = -11 \quad \text{and} \quad (-14)(3) = -42 \]

\[ 21y^2 – 14y + 3y – 2 \]

Grouping the terms:

\[ (21y^2 – 14y) + (3y – 2) \]

\[ 7y(3y – 2) + 1(3y – 2) \]

\[ (7y + 1)(3y – 2) \]

Step 3: Find the Zeros

\[ (7y + 1)(3y – 2) = 0 \]

\[ 7y + 1 = 0 \Rightarrow y = -\frac{1}{7} \]

\[ 3y – 2 = 0 \Rightarrow y = \frac{2}{3} \]

Hence, the zeros are:

\[ \alpha = -\frac{1}{7},\quad \beta = \frac{2}{3} \]

Step 4: Verify the Relationship Between Zeros and Coefficients

Given polynomial:

\[ q(y) = 7y^2 – \frac{11}{3}y – \frac{2}{3} \]

Here, \[ a = 7,\; b = -\frac{11}{3},\; c = -\frac{2}{3} \]

Sum of the Zeros

\[ \alpha + \beta = -\frac{1}{7} + \frac{2}{3} = \frac{-3 + 14}{21} = \frac{11}{21} \]

\[ -\frac{b}{a} = -\frac{-11/3}{7} = \frac{11}{21} \]

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

Product of the Zeros

\[ \alpha\beta = \left(-\frac{1}{7}\right)\left(\frac{2}{3}\right) = -\frac{2}{21} \]

\[ \frac{c}{a} = \frac{-2/3}{7} = -\frac{2}{21} \]

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

Conclusion

The zeros of the given quadratic polynomial are:

\[ -\frac{1}{7} \text{ and } \frac{2}{3} \]

The relationship between the zeros and the coefficients is verified.

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