Question

prove that( 3-√7)whole square is irrational number​

Answer 1

To Prove: (3 - √7)^2 is irrational

Solution:

${(3 - \sqrt{7} )}^{2} \\ by \: using \: {(a - b)}^{2} = {a}^{2} - 2ab + {b}^{2} \\ {(3 - \sqrt{7} )}^{2} \\ = {(3)}^{2} - (2 \times 3 \times \sqrt{7} ) + { (\sqrt{7}) }^{2} \\ = 9 - 6 \sqrt{7} + 7 \\ = 16 - 6 \sqrt{7}$

So, we got the solution which contains √7. As, √7 is irrational the solution of (3-√7)^2 is irrational.

Hope It Helps