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(x2 + 1)2 - x2 = 0 has

  • a)
    no real roots

  • b)
    4 real roots

  • c)
    2 real roots

  • d)
    1 real root

Correct answer is option 'A'. Can you explain this answer?
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**Solution:**

To solve the given equation, we can start by simplifying the expression on the left-hand side:

(x^2 + 1)^2 - x^2 = 0

Expanding the square of the binomial, we get:

(x^4 + 2x^2 + 1) - x^2 = 0

Combining like terms, we have:

x^4 + x^2 + 1 = 0

Now, let's substitute a variable to make it easier to solve. Let's set y = x^2. Substituting this into the equation, we get:

y^2 + y + 1 = 0

This is now a quadratic equation in terms of y. We can use the quadratic formula to solve for y:

y = (-b ± √(b^2 - 4ac)) / 2a

In this case, a = 1, b = 1, and c = 1. Substituting these values into the formula, we get:

y = (-1 ± √(1 - 4(1)(1))) / 2(1)

Simplifying further:

y = (-1 ± √(1 - 4)) / 2

y = (-1 ± √(-3)) / 2

Since the discriminant (√(-3)) is negative, the quadratic equation has no real roots. Therefore, the original equation (x^2 + 1)^2 - x^2 = 0 also has no real roots.

Hence, the correct answer is option A) no real roots.
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Community Answer
(x2 + 1)2 - x2 = 0 hasa)no real rootsb)4 real rootsc)2 real rootsd)1 r...
Option (A)

Given, 
(
x
2
+
1
)
2
x
2
=
0

(
x
2
+
1
)
2
=
x
2

Square root on both sides. We get,

x
2
+
1
=
x
o
r
x
2
+
1
=
x

x
2
x
+
1
=
0
o
r
x
2
+
x
+
1
=
0

discriminant for the above equations will be;

1-4 = -3 <0 or 1-4=-3 <0

Hence, 
(
x
2
+
1
)
2
x
2
=
0
 has no real roots.
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