Grade 10Trigonometry
The number of distinct positive real roots of equations (x^2+6) ^2-35x^2=2x(x^2+6)
The number of distinct positive real roots of equations (x^2+6) ^2-35x^2=2x(x^2+6)
2 Answers
surbhi dadhaniya
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
(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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