Question

Given that root 2 is irrational prove that ( 5+ root 2 )^2 is also irrational .

Answer 1

✪AnSwEr

${\tt{\red{\underline{\large{Given}}}}}$

• √2 is irrational no

${\sf{\green{\underline{\large{To\:prove}}}}}$

• (5+√2)² is also rational

${\sf{\pink{\underline{\Large{Proof}}}}}$

=>(5+√2)²

=>25+2+10√2

=>27+10√2

Let 27+10√2 as a rational no

Therefore

• it must be exist in a/b form

27+10√2=a/b

=>10√2={a/b}-27

=>10√2={a-27b}/b

=>√2={a-27b}/10b

Here 10b is a rational no

• Then √2 also a rational no
• But it given that √2 is irrational
• Therefore our assumption is wrong

Hence (5+√2)² is irrational no

Answer 2

$\bigstar$ Given :

• $\bold{\sqrt{2}\;\;is\;an\;irrational\;number }$

$\bigstar$ To Prove :

• $\bold{(5+\sqrt{2} )^2\;\;is\;an\;irrational\;number}$

$\bigstar$ Proof :

$\bold{(5+\sqrt{2})^2}\\\\\implies \bold{25+2+10\sqrt{2}}\\\\\implies \bold{27+10\sqrt{2}}$

Let us assume that 27+10√2 is a rational number .

$\implies \bold{27+10\sqrt{2}=\frac{a}{b} }\\\\\implies \bold{\sqrt{2}=\frac{a}{b}-27}\\\\\implies \bold{10\sqrt{2}=\frac{(a-27)}{b} }\\\\\implies \bold{\sqrt{2}=\frac{(a-27)}{10b} }$

Since it is proved that √2 is a rational number .

Thus our assumption is wrong .

So $\bold{(5+\sqrt{2})^2}$ is also irrational.