SOLUTION: Please help me solve the following problem with working: {{{0 = x^2 - (sqrt(2) + sqrt(5))x + sqrt(10)}}} My answer is.. {{{x = ((sqrt(2) + sqrt(5))+ sqrt(7 - 2sqrt(10)))/2

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Please help me solve the following problem with working: {{{0 = x^2 - (sqrt(2) + sqrt(5))x + sqrt(10)}}} My answer is.. {{{x = ((sqrt(2) + sqrt(5))+ sqrt(7 - 2sqrt(10)))/2      Log On


   

Question 714958: Please help me solve the following problem with working:
0+=+x%5E2+-+%28sqrt%282%29+%2B+sqrt%285%29%29x+%2B+sqrt%2810%29
My answer is..
x+=+%28%28sqrt%282%29+%2B+sqrt%285%29%29%2B+sqrt%287+-+2sqrt%2810%29%29%29%2F2
The answer is apparently sqrt%282%29 and sqrt%285%29.
Cheers,
Dio
Found 3 solutions by jim_thompson5910, lynnlo, Alan3354:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
There's probably some trick to solving or simplifying nested radicals, but it's not coming to mind at the moment.

Anyways, there's a much simpler way to see why the answers are sqrt%282%29 and sqrt%285%29

Rewrite the last term sqrt%2810%29 as

sqrt%2810%29=sqrt%282%2A5%29

sqrt%2810%29=sqrt%282%29%2Asqrt%285%29

Now use the idea that if the roots of x%5E2+-+bx%2Bc are p and q, then b+=+p%2Bq and c+=+p%2Aq

So if you replace p with sqrt%282%29 and q with sqrt%285%29, you get

b+=+sqrt%282%29%2Bsqrt%285%29 and q+=+sqrt%282%29%2Asqrt%285%29


Hopefully that makes sense to you.

Answer by lynnlo(4176) About Me  (Show Source):
You can put this solution on YOUR website!
what are you asking your answer to your problem is correct
√2,√5

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Please help me solve the following problem with working:

0+=+x%5E2+-+%28sqrt%282%29+%2B+sqrt%285%29%29x+%2B+sqrt%2810%29
---------------
It can be factored.
%28x+-+sqrt%282%29%29%2A%28x+-+sqrt%285%29%29+=+0
x+=+sqrt%282%29
x+=+sqrt%285%29