Question

show that (3-√5)^2 is irrational​

Answer 1

Answer:

Step-by-step explanation:

On solving (3-√5)^2

                  by using identity ; (a-b)^2 =a^2+b^2-2ab

                (3-√5)^2 = 9+5-2*3*(-√5)

                                =14-(-6√5)

                                =14+6√5                        ......1eq.

let,14+6√5 is rational no.

On putting root 5 aside,and putting 1 eq. equal to 0,

then,

                               14+6√5 = 0

                                     6√5 = -14

                                       √5 = -14/6

                                       √5 = -7/3                            (since,√5=irrational no.

                                                                                   and -7/3 = rational no.)

                                    Irrational no.=Rational no.

              which is not possible,so,it contradicts our assumption that 14+6√5 is rational no.

hence ,14+6√5 is irrarional no. .i.e. (3-√5)^2 is irrational no.

HOPE THIS HELPS....