The number of real solutions of the equation 3( x² + 1/x² ) - 2( x + 1/x ) + 5 = 0, is

Solution

Maths sir · 5 steps

Put t = x + 1/x, then check if those t values are even possible for a real x.

1. Substitute. Let t = x + 1/x. Squaring gives t² = x² + 2 + 1/x², so x² + 1/x² = t² - 2.

   3(t² - 2) - 2t + 5 = 0 ⇒ 3t² - 2t - 1 = 0 ⇒ (3t + 1)(t - 1) = 0

2. Values of t. So t = 1 or t = -1/3.
3. Range of t. For real x > 0, AM–GM gives x + 1/x ≥ 2. For real x < 0, it gives x + 1/x ≤ -2. So a real x always gives |t| ≥ 2, and neither 1 nor -1/3 is allowed.
4. Cross-check. With t = 1 you get x² - x + 1 = 0, where D = 1 - 4 = -3. With t = -1/3 you get 3x² + x + 3 = 0, where D = 1 - 36 = -35. Both discriminants are negative, so neither equation has a real root.
5. So. The equation has 0 real solutions, which is option (c).

The idea to keep. Whenever you substitute t = x + 1/x, remember the restriction |t| ≥ 2 for real x. Throw out any t value outside that range before you count solutions.

Answer from the NTA final answer key. Solution by PrepWiser.