Question

Find the zeros of the following quadratic polynomial and verify the relationship between the zeros and the coefficients:

`phi(x) = 2x^2 + 7/2x + 3/4`

Answer 1

Given: `phi(x) = 2x^2 + 7/2x + 3/4`

Step-wise calculation:

1. Identify coefficients:

`a = 2, b = 7/2, c = 3/4`

2. Discriminant: Δ = b² – 4ac 

= `(7/2)^2 - 4 xx 2 xx (3/4)` 

= `49/4 - 6`

= `25/4`

`sqrt(Δ) = 5/2`

3. Roots (quadratic formula):

`x = (-b ± sqrt(Δ))/(2a)`

= `(-(7/2) ± (5/2))/4`

= `(-7 ± 5)/8` 

Hence the zeros are `x_1 = (-7 + 5)/8`

= `(-2)/8`

= `(-1)/4` 

And `x_2 = (-7 - 5)/8`

= `(-12)/8`

= `(-3)/2`

4. Verify relationship between zeros and coefficients.

For a quadratic `ax^2 + bx + c, α + β = -b/a` and `αβ = c/a`.

Sum: `α + β = (-1/4) + (-3/2)`

= `-1/4 - 6/4`

= `-7/4`

`-b/a = -(7/2)/2`

= `-7/4`

Product: `αβ = (-1/4) xx (-3/2) = 3/8`

`c/a = (3/4)/2`

= `3/8`

The zeros of `phi(x)` are `x = −1/4` and `x = −3/2` and they satisfy `α + β = -b/a` and `αβ = c/a` relationship verified.