Zeros of a Quadratic Polynomial

Video Explanation

Question

Find the zeros of the quadratic polynomial

\[ h(t) = t^2 – 15 \]

and verify the relationship between the zeros and their coefficients.

Solution

Step 1: Find the Zeros of the Polynomial

Given:

\[ h(t) = t^2 – 15 \]

Factorising:

\[ t^2 – 15 = (t – \sqrt{15})(t + \sqrt{15}) \]

Equating each factor to zero:

\[ t – \sqrt{15} = 0 \Rightarrow t = \sqrt{15} \]

\[ t + \sqrt{15} = 0 \Rightarrow t = -\sqrt{15} \]

Hence, the zeros of the polynomial are:

\[ \alpha = \sqrt{15},\quad \beta = -\sqrt{15} \]

Step 2: Verify the Relationship Between Zeros and Coefficients

For a quadratic polynomial \[ at^2 + bt + c, \]

Sum of zeros:

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

Product of zeros:

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

Here, \[ a = 1,\; b = 0,\; c = -15 \]

Sum of the zeros

\[ \alpha + \beta = \sqrt{15} + (-\sqrt{15}) = 0 \]

\[ -\frac{b}{a} = -\frac{0}{1} = 0 \]

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

Product of the zeros

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

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

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

Conclusion

The zeros of the given quadratic polynomial are:

\[ \sqrt{15} \text{ and } -\sqrt{15} \]

The relationship between the zeros and the coefficients is verified.

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