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The number of distinct positive real roots of equations (x^2+6) ^2-35x^2=2x(x^2+6)

Nagarjunareddy , 3 Years ago
Grade 10
anser 2 Answers
surbhi dadhaniya

Last Activity: 3 Years ago

divide whole equation by x^2
gives: [(x2+6)/x]2-35=2[(x2+6)/x]
now let (x2+6)/x=t
t2-2t-35..... solve this simple quadratic
give t=5 and t=-7
equate these with (x2+6)/x one by one
solving we get x=2 x=3 x=-6 x=-1
it has got 2 distinct positive real roots namely 2 and 3

Somya

Last Activity: 3 Years ago

(X²+6) ² - 35x² = 2x(x²+6) 
=>[(x²+6)/X]² - 35 = 2( x²+6) /x  {divide by x²)}
t² - 35 = 2t ( assuming (x²+6) /x as a variable t) 
(t-7) (t+5) = 0 
Now solving for x we get 
(X²+6)/x = 7 Or -5 and hence x = 1,6,-2, -3 are answers. 

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