Question

(x² + 1)² – x² = 0 has ______.

Options

• Four real roots

• Two real roots

• No real roots

• One real root

Answer 1

(x² + 1)² – x² = 0 has no real roots.

Explanation:

Given equation is `(x^2 + 1)^2 - x^2` = 0

⇒ `x^4 + 1 + 2x^2 - x^2 = 0`  ......`[because (a + b)^2 = a^2 + b^2 + 2ab]`

⇒ `x^4 + x^2 + 1` = 0

Let `x^2 = y`

∴ `(x^2)^2 + x^2 + 1` = 0

`y^2 + y + 1` = 0

On comparing with `ay^2 + 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

∴ `y^2 + y + 1` = 0

i.e., `x^4 + x^2 + 1` = 0

or `(x^2 + 1)^2 - x^2` = 0 has no real roots.