prove that root 11 + root 7 is an irrational
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√11+√7
spoz a= √11+√7
a= √11+√7
a-√11=√7
(a-√11)^2=(√7)^2
a^2-2√11a+11^2=49
a^2 -2√11a+121 = 49
a^2-2√11a+121-49=0
a^2-2√11a+72=0
but √11 is irrational number
so √11+√7 is an irrational number..
this is your answer....
hope it helps you....
spoz a= √11+√7
a= √11+√7
a-√11=√7
(a-√11)^2=(√7)^2
a^2-2√11a+11^2=49
a^2 -2√11a+121 = 49
a^2-2√11a+121-49=0
a^2-2√11a+72=0
but √11 is irrational number
so √11+√7 is an irrational number..
this is your answer....
hope it helps you....
Naman3102:
Thank a lot to you bro
how (√7)^2 = 49
Answered by
3
Answer:
Step-by-step explanation:
√11+√7
spoz a= √11+√7
a= √11+√7
a-√11=√7
(a-√11)^2=(√7)^2
a^2-2√11a+11^2=49
a^2 -2√11a+121 = 49
a^2-2√11a+121-49=0
a^2-2√11a+72=0
but √11 is irrational number
so √11+√7 is an irrational number..
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