Question

The harmonic mean of the roots of the equation $ (5 + \sqrt 2 ){x^2} - (4 + \sqrt 5 )x + 8 + 2\sqrt 5 = 0 $ .
 $
  A.2 \\
  B.4 \\
  C.6 \\
  D.8 \\
 $

Answer 1

Hint: In order to get the right answer we need to use the formula of harmonic mean for two numbers and calculate the sum and product of roots of the equation given and then put those values in the formula of Harmonic mean. Doing this will solve your problem and will give you the right answer.

Complete step-by-step answer:
The given equation is $ (5 + \sqrt 2 ){x^2} - (4 + \sqrt 5 )x + 8 + 2\sqrt 5 = 0 $ ……….(1)
It is a quadratic equation that will have two roots.
We know that the harmonic mean of two numbers p and q will be HM = $ \dfrac{{{\text{2pq}}}}{{{\text{p + q}}}} $ ………..(2)
We also know that if there is a quadratic equation $ a{x^2} + bx + c = 0 $ .
Then its sum of roots will be $ - \dfrac{b}{a} $ and product of roots will be $ \dfrac{c}{a} $ .
Let us consider p and q as the roots of the equation (1)
Then sum of roots will be p + q = $ \dfrac{{4 + \sqrt 5 }}{{5 + \sqrt 2 }} $ and product of root is pq = $ \dfrac{{8 + 2\sqrt 5 }}{{5 + \sqrt 2 }} $ .

Putting the values of sum and product of p and q in equation (2) we get harmonic mean
HM = $ \dfrac{{2\left( {\dfrac{{8 + 2\sqrt 5 }}{{5 + \sqrt 2 }}} \right)}}{{\dfrac{{4 + \sqrt 5 }}{{5 + \sqrt 2 }}}} = \dfrac{{2\left( {8 + 2\sqrt 5 } \right)}}{{4 + \sqrt 5 }} = 2\left( 2 \right) = 4 $ .
Hence the answer is 4.
So, the correct option is B.

Note:To get such problems solved you need to remember the formula of harmonic mean of two numbers. Here don’t try to find the roots of the equation it will take time and here it will be complex. Here you have to observe the formula of harmonic mean which is contained in the sum and product of roots. So, you have to obtain the sum and product of roots of the quadratic equation and put those values in the formula of harmonic mean to get the answer of harmonic mean. Remembering the technique to obtain the sum and product of quadratic, cubic, biquadratic etc. equations will help you further.