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Need help with this surds questions

Question: Express 5+42542 \sqrt{\frac{5+ 4 \sqrt{2}}{5- 4 \sqrt{2}}}
in the form of a+b2 a + b\sqrt{2}
where a, b ϵQ \epsilon Q


5+42542 \sqrt{\frac{5+ 4 \sqrt{2}}{5- 4 \sqrt{2}}}

=(5+4254200.5 =(\frac{5+ 4 \sqrt{2}}{5- 4 \sqrt{2}}0^ {0.5}

5+42542 \sqrt{\frac{5+ 4 \sqrt{2}}{5- 4 \sqrt{2}}}

((5+42)(5+42))((5+42)(542)) \sqrt{\frac{((5 +4\sqrt{2}) (5 +4\sqrt{2}))}{((5 +4\sqrt{2}) (5 -4\sqrt{2}))}}

25+202+202+(16(2))2532 \sqrt{\frac{25+20 \sqrt{2}+ 20 \sqrt{2}+(16(2))}{25-32}}

25+202+202+(16(2))7 \sqrt{\frac{25+20 \sqrt{2}+ 20 \sqrt{2}+(16(2))}{-7}}


Please I need help from here what do i do?
Reply 1
Original post by bigmansouf

Please I need help from here what do i do?


You are screwed. Notice that the insides of the square root are negative, how are you going to get a real number?
Reply 2
Original post by GPiph
You are screwed. Notice that the insides of the square root are negative, how are you going to get a real number?


did i do anything wrong
could i have approach the question differently
Reply 3
Original post by bigmansouf
did i do anything wrong
could i have approach the question differently


Are you sure you have quoted the question correctly?
The problem you have given can't be solved.
Reply 4
q14.jpg
Original post by GPiph
Are you sure you have quoted the question correctly?
The problem you have given can't be solved.
Reply 5
Original post by bigmansouf
q14.jpg


As I said, clearly 5425 - 4 \sqrt{2} is negative, and 5+425+4\sqrt{2} is positive, so 5+42542\dfrac{5+4\sqrt{2}}{5 - 4 \sqrt{2}} is negative. Suppose it can be written as a+b2=5+42542\displaystyle a+b\sqrt{2} = \sqrt{\dfrac{5+4\sqrt{2}}{5 - 4 \sqrt{2}}}, then (a+b2)2=5+42542\displaystyle (a+b\sqrt{2})^2 = \dfrac{5+4\sqrt{2}}{5 - 4 \sqrt{2}}, the RHS being negative but the LHS being non-negative. Contradiction. So it can't be written in that form.
Reply 6
Original post by GPiph
As I said, clearly 5425 - 4 \sqrt{2} is negative, and 5+425+4\sqrt{2} is positive, so 5+42542\dfrac{5+4\sqrt{2}}{5 - 4 \sqrt{2}} is negative. Suppose it can be written as a+b2=5+42542\displaystyle a+b\sqrt{2} = \sqrt{\dfrac{5+4\sqrt{2}}{5 - 4 \sqrt{2}}}, then (a+b2)2=5+42542\displaystyle (a+b\sqrt{2})^2 = \dfrac{5+4\sqrt{2}}{5 - 4 \sqrt{2}}, the RHS being negative but the LHS being non-negative. Contradiction. So it can't be written in that form.

thank you very much for taking your time to answer
i was struggling with this

so just to confirm the question was worded wrong
Reply 7
Original post by bigmansouf
thank you very much for taking your time to answer
i was struggling with this

so just to confirm the question was worded wrong


Yep.
Reply 8
Original post by bigmansouf
thank you very much for taking your time to answer
i was struggling with this

so just to confirm the question was worded wrong


Rah, ur teacher must be dodo ak
Reply 9
Original post by Naruke
Rah, ur teacher must be dodo ak


lool the question is from crashmaths
Reply 10
Original post by bigmansouf
lool the question is from crashmaths


Have a look at madasmaths, if you want questions that actually work...
Original post by Naruke
Have a look at madasmaths, if you want questions that actually work...


lool thanks chief :biggrin:
Original post by Naruke
Have a look at madasmaths, if you want questions that actually work...


Please can you check this question for me if it is the same wrong format question
Original post by Naruke
Have a look at madasmaths, if you want questions that actually work...


q11.jpg
my solution:
i up[loaded my solution
am right that there is something wrong?
Original post by bigmansouf
q11.jpg
my solution:
i up[loaded my solution
am right that there is something wrong?


Yeah you're right but obviously the second equation is defined for x,y0,xy0\forall x,y\geq 0, \sqrt{x} - \sqrt{y}\not= 0 so they never intersect. Unless the domain of both equations is over C\mathbb{C}
(edited 7 years ago)
Original post by RDKGames
Yeah you're right but obviously the second equation is defined for x,y0,xy0\forall x,y\geq 0, \sqrt{x} - \sqrt{y}\not= 0 so they never intersect. Unless the domain of both equations is over C\mathbb{C}


Thank RDKGames hows Maths at uni?
Original post by bigmansouf
Thank RDKGames hows Maths at uni?


No problem. :smile:

It's great imo, we are flying through content quite fast so it's a challenge to keep up to be honest with everything else going on at uni socially xD but content-wise it's great, learning loads of new material that is barely even scratched at A-Level - FM or not :smile:
Original post by RDKGames
No problem. :smile:

It's great imo, we are flying through content quite fast so it's a challenge to keep up to be honest with everything else going on at uni socially xD but content-wise it's great, learning loads of new material that is barely even scratched at A-Level - FM or not :smile:


cool
Original post by Naruke
Have a look at madasmaths, if you want questions that actually work...


Unfortunately, I produce a lot of questions and sometimes errors arise! There are no solutions to these questions and so that is why they were undetected.

I apologise for the errors, I'll rectify them when I have time


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