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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: scientificllama on June 05, 2019, 01:09:16 am

Title: Midnight Maths Problems
Post by: scientificllama on June 05, 2019, 01:09:16 am
It's that time again, exam season, and I have failed to miss 3 questions that I have no clue to do. Could someone help me out? Thanks in advance ;)
Title: Re: Midnight Maths Problems
Post by: AlphaZero on June 05, 2019, 01:20:42 am
It's that time again, exam season, and I have failed to miss 3 questions that I have no clue to do. Could someone help me out? Thanks in advance ;)

I'll start you off.

Q1:  You should have found that \((\sqrt{3}+1)^2=4+2\sqrt{3}\), so \[\sqrt{16+8\sqrt{3}}=\sqrt{4(4+2\sqrt{3})}=\dots\]

Q2:\[\sqrt{(5-2)^2+(y+1)^2}=5\implies \dots\]

Q3:\[(2x-3)(4x^2+6x+9)=8x^3+0x^2+0x-27\]
Title: Re: Midnight Maths Problems
Post by: scientificllama on June 05, 2019, 01:28:37 am
I'll start you off.

Q1:  You should have found that \((\sqrt{3}+1)^2=4+2\sqrt{3}\), so \[\sqrt{16+8\sqrt{3}}=\sqrt{4(4+2\sqrt{3})}=\dots\]

Q2:\[\sqrt{(5-2)^2+(y+1)^2}=5\implies \dots\]

Q3:\[(2x-3)(4x^2+6x+9)=8x^3+0x^2+0x-27\]

Wow, didn't expect such a quick reply. Thanks for the help. To be honest, I have no actual idea of how to do questions 1 and 3... :-\