Get the solution to "Suppose a2−4b≠0. Let α, β be the (distinct)..."

Javion Kerr

Answered question

2022-04-03

Suppose

a^{2} - 4 b \neq 0

. Let

α, β

be the (distinct) roots of the polynomial

x^{2} + a x + b

Then there is a real number c s.t.

kachnaemra

Beginner2022-04-04Added 16 answers

Step 1
Your proof of case 2 is clearly incomplete. Note that, in that case,

α = \frac{- a \pm \sqrt{a^{2} - 4 b}}{2} and β = \frac{- a \mp \sqrt{a^{2} - 4 b}}{2} .

Since

a^{2} - 4 b < 0

, this means that

α = \frac{- a \pm \sqrt{4 b - a^{2}}, i}{2} and that β = \frac{- a \mp \sqrt{4 b - a^{2}}, i}{2} .

So,

α - β = \pm \sqrt{4 b - a^{2}}, i .

sorrisi7yny

Beginner2022-04-05Added 9 answers

Step 1

{(α - β)}^{2} = {(α + β)}^{2} - 4 α β = a^{2} - 4 b \in R

a^{2} - 4 b > 0

, then

α - β = c

where

c^{2} = a^{2} - 4 b

a^{2} - 4 b < 0

, then

α - β = c i

, where

c^{2} = - (a^{2} - 4 b)

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?