Math, asked by nidaynf, 10 months ago

If (a=3-√5/3+√5),(b=3+√5/3-√5). Find the value of (a^2+b^2).

Answers

Answered by sunita926343
2

Step-by-step explanation:

a=3-√5/3+√5

a=3-√5(3-√5)/(3)²-(√5)²

a=9-3√5-3√5+5/9-5

a=14-6√5/4

b=3+√5/3-√5

b=3+√5(3+√5)/(3)²-(√5)²

b=9+3√5+3√5+5/9-5

b=14+6√5/4

a²-b²=(a+b)(a-b)

        =(14-6√5/4+14+6√5/4)(14-6√5/4-14+6√5/4)

        =(14-6√5+14+6√5/4)(14-6√5-14+6√5/4)

        =28/4*0/4

        =7*0

        =0

therefore the value is 0

hope it helps

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