Question

please answer step by step all parts of the question for 25 marks​

Answer 1

1)

$\sqrt{5}$

is an irrational number because it cannot be written in the p/q form.

2)

$\pi$

pie also cannot be written in the p/q form so it is an irrational number.

3)

$\sqrt{49}$

square root of 49 can also be written as 7 so seven can be written in the p/q form i.e, 7/1 hence it is a rational number.

4)

$3 + \sqrt{7}$

just assume that 3 +root 7 is rational

so ,

3 + root 7 = p/q

root 7 = p/q - 3/1 {Cross multiplication}

root 7 = 3b - a / b -------------- eqn 1

Here 3b, a and b are integers and the whole eqn 1 is rational which is equal to an irrational number i.e, root 7. since root 7 is irrational but according to this it shows rational we have to change our assumption.

5) (4+ root 5)^2

we can use one identity to expand this...
a + b the whole square .
according to the question,
=(4 + root 5 )^2
=(4^2) + (2 × 4 × root 5) + (root =5^2)

=16 + 8root 5 + 5
=21 + 8root 5
=38.88854382......
which is a non terminating decimal so it is irrational.

6) ( 3 + root 7)( 3 - root 7)
this also is in a form of an algerbraic equation i.e , (a+b)(a-b) which is equal to a^2 - b^2 so, according to the question

=(3 + root 7) (3 - root 7)
=3^2 - root 7 ^2
=9 - 7
=2

so 2 can be written as p/q form so given equation is a rational.

Ignore my flaws.
i hope that answers your question!

Answer 2

i) irrational

ii) irrational

iii) rational

iv) irrational

v) irrational

Vi) rational