Question

√(13-x√10= √8+ √5 ) then what is the value of x ?

Answer 1

√(13-x√10= √8+ √5 ) then what is the value of x?
   √13-x√10= √8+ √5 
squaring on both sides (a+b)^2 = a^2 + b^2 + 2ab
  (√13)^2 + (x√10)^2 - 2 x√130 = (√8)^2 + (√5)^2 + 2√40
    13 + 10 x^2 - 2x√130   = 8 + 5 + 2√40   (sub both side by 13)
10x^2 - 2x√130  =  2√40  (divide both side by 2)
5x^2 - x√130 = √40

5x^2 - x√130 = √40=( -b ­+ (√b2 – 4ac)) / 2a

                               = (√130  + (130 – 4x5x√40)) /2x5

                               = √130/10 + (13 - 2√40)

                           

Answer 2

By squaring both side we get
13-x√10=8+5+2√8√5;

13-x√10=13+2√40 cancel 13 from both side we get,
-x√10=2*2√10
now cancel √10 from both side
we get x=-4