Question

Obtain the zeroes of the polynomial 7x² + 18x – 9. Hence, write a polynomial each of whose zeroes is twice the zeroes of the given polynomial.

Answer 1

Given: 7x² + 18x – 9.

Step-wise calculation:

1. Discriminant: Δ = b² – 4ac 

= `18^2 - 4 xx 7 xx (-9)` 

= 324 + 252

= 576

2. `sqrt(Δ) = 24`.

3. Roots: `x = (-18 ± 24)/(2 xx 7)`

`x_1 = (-18 + 24)/14`

= `6/14`

= `3/7`

`x_2 = (-18 - 24)/14`

= `(-42)/14`

= –3

4. Double each zero:

`2x_1 = 6/7`

2x₂ = –6

5. Sum of new zeros S' = `6/7 + (-6) = −36/7`.

Product of new zeros P' = `(6/7) xx (-6) = -36/7`.

6. Required monic polynomial:

`x^2 - S'x + P' = x^2 + (36/7)x - 36/7`

Clearing denominators (multiply by 7): 7x² + 36x – 36.

Zeroes of 7x² + 18x – 9 are `3/7` and –3.

A polynomial whose zeroes are twice those is 7x² + 36x – 36.