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(x2+1)2−x2 = 0 has
- a)1 real root
- b)2 real roots
- c)4 real roots
- d)no real roots
Correct answer is 'A'. Can you explain this answer?
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Verified Answer
(x2+1)2−x2= 0 hasa)1 real rootb)2 real rootsc)4 real rootsd)no r...
Given: (x2+1)2 − x2 = 0
⇒ x2 + 1 + 2x − x2 = 0
⇒ 2x + 1 = 0
⇒ x = −1/2
Therefore, (x2+1)2 − x2 = 0 has no real roots.
⇒ x2 + 1 + 2x − x2 = 0
⇒ 2x + 1 = 0
⇒ x = −1/2
Therefore, (x2+1)2 − x2 = 0 has no real roots.
Most Upvoted Answer
(x2+1)2−x2= 0 hasa)1 real rootb)2 real rootsc)4 real rootsd)no r...
(x² + 1)² = x²
x² + 1 = ± x
• x² + 1 = x
x² - x + 1 = 0
Since the discriminant is less than zero there will be no real roots.
• x² + 1 = -x
x² + x + 1 = 0
Since the discriminant is less than zero there will be no real roots.
So I think there will be no real roots possible.
Another way to do is that since square terms are always greater than zero (x² + 1)² + x² = 0 is satisfied only when both x² + 1 = 0 and x² = 0 and there is no real value of x which will satisfy both.
x² + 1 = ± x
• x² + 1 = x
x² - x + 1 = 0
Since the discriminant is less than zero there will be no real roots.
• x² + 1 = -x
x² + x + 1 = 0
Since the discriminant is less than zero there will be no real roots.
So I think there will be no real roots possible.
Another way to do is that since square terms are always greater than zero (x² + 1)² + x² = 0 is satisfied only when both x² + 1 = 0 and x² = 0 and there is no real value of x which will satisfy both.
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Community Answer
(x2+1)2−x2= 0 hasa)1 real rootb)2 real rootsc)4 real rootsd)no r...
It will have 4 real roots because highest degree is of 4 (x^2+1)2-x^2=0 therefore, x^4+x^2+2=0
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