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)
things were not as complicated when I was in middle school.
Need a photo of the graph to help.
Definition of eo-primes or relatively primes: Two numbers are said to be co-prime or relatively prime If their HCF IS 1 Hence to prove 847 and 2160 as co-prime numbers we will find their HCF and which should be 1
New steps to find HCF will be as under
2160 = 847 x 2+ 466
847 = 466 ×1 +381
466 = 381 x 1 + 85
381 =85 x 4+ 41
85 =41 x 2+3
41 =3 x 13+ 2
3 =2 x 1+1
2 =1 x 2+0
Therefore, the HCF=1 Hence, the numbers are co-primes (relatively prime).