Zeros of a Quadratic Polynomial

Video Explanation

Question

Find the zeros of the quadratic polynomial

\[ g(x) = a(x^2 + 1) – x(a^2 + 1) \]

and verify the relationship between the zeros and their coefficients.

Solution

Step 1: Write the Polynomial in Standard Form

\[ g(x) = ax^2 + a – (a^2 + 1)x \]

\[ g(x) = ax^2 – (a^2 + 1)x + a \]

Step 2: Factorise the Polynomial

\[ ax^2 – (a^2 + 1)x + a \]

Rewriting:

\[ ax^2 – a^2x – x + a \]

Grouping the terms:

\[ (ax^2 – a^2x) – (x – a) \]

\[ ax(x – a) – 1(x – a) \]

\[ (ax – 1)(x – a) \]

Step 3: Find the Zeros

\[ (ax – 1)(x – a) = 0 \]

\[ ax – 1 = 0 \Rightarrow x = \frac{1}{a} \]

\[ x – a = 0 \Rightarrow x = a \]

Hence, the zeros are:

\[ \alpha = a,\quad \beta = \frac{1}{a} \]

Step 4: Verify the Relationship Between Zeros and Coefficients

For a quadratic polynomial \(Ax^2 + Bx + C\):

\[ \alpha + \beta = -\frac{B}{A},\quad \alpha\beta = \frac{C}{A} \]

Here, \[ A = a,\; B = -(a^2 + 1),\; C = a \]

Sum of the Zeros

\[ \alpha + \beta = a + \frac{1}{a} = \frac{a^2 + 1}{a} \]

\[ -\frac{B}{A} = -\frac{-(a^2 + 1)}{a} = \frac{a^2 + 1}{a} \]

\[ \alpha + \beta = -\frac{B}{A} \quad \checkmark \]

Product of the Zeros

\[ \alpha\beta = a \times \frac{1}{a} = 1 \]

\[ \frac{C}{A} = \frac{a}{a} = 1 \]

\[ \alpha\beta = \frac{C}{A} \quad \checkmark \]

Conclusion

The zeros of the given quadratic polynomial are:

\[ a \text{ and } \frac{1}{a} \]

The relationship between the zeros and the coefficients is verified.

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