Question

4.
k+1k+2
If the roots of ax2+bx+c = 0 are
k 'k+1
then (a+b+c)2 =
1) 0
4) 2abc
2) 2
3) 62-4ac​

Answer 1

Step-by-step explanation:

Note that a≠0 since ax2+bx+c=0 has two ( distinct ) roots. So, if α , β denote the roots of ax2+bx+c=0 , then

(a+b+c)2=b2–4ac⇔(ca+ba+1)2=(ba)2−4ca

⇔(αβ−(α+β)+1)2=(α+β)2−4αβ=(α−β)2

⇔|(α−1)(β−1)|=|α−β|…(⋆)

With α=1+1k and β=1+1k+1 , (⋆) reduces to the verification

∣∣1k(k+1)∣∣=∣∣1k−1k+1∣∣ .

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