Question

2 square root 5+square root 7 prove that it is irrational number
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Answer 1

Answer:

Step-by-step explanation:

2square root 5 + square root 7

Let us assume a contradiction that 2square root 5+square root 7 is rational number

2square root 5+square root 7 = p/q where p and q are co-primes and q is not equal to 0

Squaring on both sides

(2square root 5+ square root 7 )^2 = (p/q)^2

We get

20+7+4square root 35= (p/q)^2

27+4square root 35 =(p/q)^2

4square root 35 = (p^2/q^2)-27

4square root 35= p^2-27q^2/q^2

Square root 35 = p^2-27q^2/4q^2

Where p And q are integers and p^2-27q^2/4q^2 is rational number

So root 35 is rational number

But our assumption is wrong

Root 35 is irrational number

So 2root5+root 7 is irrational number

Hence proved

Answer 2

$\huge\boxed{Answer:-}$

√35 is rational number

so, 2√5+√7 is irrational number..

Step-by-step explanation:

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