3) x² - 121 = 0
x² = 121
x' = +√121
x' = 11
_______________
x'' = - √121
x'' = -11
Solution ⇒ S{-11 ; 11 } or (x-11)(x+11)
4) 4x² + 144 = 0
4x² = -144
x² = -144 / 4
x² = -36
x = √-36
No solution ⇒ S = ∅
5) z²+10z+21 = 0
Δ = 10² - 4(1)(21)
Δ = 100 - 84
Δ = 16
x' = (-10+4) / 2 = -6/2 = -3
x'' = (-10-4) / 2 = -14/2 = -7
Solution ⇒ S{ -7 ; -3} or (x+3)(x+7)
Answer:
A)29 and B) 21,
Step-by-step explanation:
First,in the sequence , the parameter must to be a integer.
Second, we need to solve the equations by .
All the option in the problem represent an
Then, we need to prove all number in the options, if the result is a integer number, this option can be part of the sequence.
For A)
For B)
For C)
For D)
Only A) and B) only A and B meet the requirement
You are correct. here it is step by step.
[7-(2-5(2)+7)] +2
[7-(2-10+7)]+2
[7-(-1)]+2
8+2
10
Answer:
1= 2x+11
2= is none of those
3= √x^2 –11
4= √2x+11
Step-by-step explanation: