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(x2 + 1)2 - x2 = 0 has

  • a)
    Four real roots

  • b)
    Two real roots

  • c)
    No real roots

  • d)
    One real root

Correct answer is option 'C'. Can you explain this answer?
Verified Answer
(x2 + 1)2 - x2 = 0 hasa)Four real rootsb)Two real rootsc)No real roots...
Given equation is (x2 + 1)2 - x2 = 0
⇒ x4 + 1 + 2x2 - x2 = 0 [∵ (a + b)2 = a2 + b2 + 2ab]
⇒ x4 + x2 + 1 = 0
Let x2 = y
∴ (x2)2 + x2 + 1 = 0
y2 + y + 1 = 0
On comparing with ay2 + by + c = 0, we get
a = 1, b = 1 and c = 1
Discriminant, D = b2 - 4ac
= (1)- 4(1)(1)
= 1 - 4 = -3
Since, D < 0
∴ y2 + y + 1 = 0 i.e., x4 + x2 + 1 = 0 or (x2 + 1)2 - x2 = 0 has no real roots.
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Most Upvoted Answer
(x2 + 1)2 - x2 = 0 hasa)Four real rootsb)Two real rootsc)No real roots...
Given equation is (x2 + 1)2 - x2 = 0
⇒ x4 + 1 + 2x2 - x2 = 0 [∵ (a + b)2 = a2 + b2 + 2ab]
⇒ x4 + x2 + 1 = 0
Let x2 = y
∴ (x2)2 + x2 + 1 = 0
y2 + y + 1 = 0
On comparing with ay2 + by + c = 0, we get
a = 1, b = 1 and c = 1
Discriminant, D = b2 - 4ac
= (1)- 4(1)(1)
= 1 - 4 = -3
Since, D < 0
∴ y2 + y + 1 = 0 i.e., x4 + x2 + 1 = 0 or (x2 + 1)2 - x2 = 0 has no real roots.
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(x2 + 1)2 - x2 = 0 hasa)Four real rootsb)Two real rootsc)No real roots...
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