Class 10 Exam > Class 10 Questions > (x2 + 1)2 - x2 = 0 hasa)Four real rootsb)Two ...
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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?
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(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)2 - 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.
⇒ 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)2 - 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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(x2 + 1)2 - x2 = 0 hasa)Four real rootsb)Two real rootsc)No real roots...
(x^2+1)^2-x^2=0,
√(x^2+1)^2=(√x^2),
x^2+1=x,
x^2-x+1=0,
Discriminant=b^2-4ac,
=(-1)^2-4(1)(1),
=1-4,
=-3,
as we know that when discriminant (D) is less than 0,
no real roots exists .
Hence the correct answer is option C
√(x^2+1)^2=(√x^2),
x^2+1=x,
x^2-x+1=0,
Discriminant=b^2-4ac,
=(-1)^2-4(1)(1),
=1-4,
=-3,
as we know that when discriminant (D) is less than 0,
no real roots exists .
Hence the correct answer is option C
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