Flag Trigonometry> The number of distinct positive real root...
question mark

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. 

Provide a better Answer & Earn Cool Goodies

Enter text here...
star
LIVE ONLINE CLASSES

Prepraring for the competition made easy just by live online class.

tv

Full Live Access

material

Study Material

removal

Live Doubts Solving

assignment

Daily Class Assignments


Ask a Doubt

Get your questions answered by the expert for free

Enter text here...