Question

Equation : (x² + 1)² - x² = 0 has real _________ root(s)​

Answer 1

Given,

f(x): $(x^2+1)^2 - x^2 = 0$

To find,

The real roots of f(x).

Solution,

The equation $(x^2+1)^2 - x^2 = 0$ has no real roots.

We can simply solve the mathematical problem by the following procedure.

We know that,

⇒ $(x^2+1)^2 - x^2 = 0$

⇒ $x^4 + 1 + 2x^2 - x^2 = 0$

⇒ $x^4 + x^2 + 1 = 0$

Let x² be m.

⇒ m² + m + 1 = 0

We know that the above-given equation does not have any real root as D< 0.

As a result, we can conclude that the equation given has no real roots.

Answer 2

(x² + 1)² – x² = 0 has real no roots

Given :

The equation (x² + 1)² – x² = 0

To find :

(x² + 1)² – x² = 0 has real _________ root(s)

Solution :

Step 1 of 4 :

Write down the given equation

The given equation is

(x² + 1)² – x² = 0

Step 2 of 4 :

Simplify the given equation

(x² + 1)² - x² = 0

⇒ x⁴ + 2x² + 1 - x² = 0

⇒ x⁴ + x² + 1 = 0

Step 3 of 4 :

Find Discriminant of the equation

Let y = x²

Then above equation becomes

x⁴ + x² + 1 = 0

⇒ y² + y + 1 = 0

Comparing with the general equation

ay² + by + c = 0 we get

a = 1 , b = 1 and c = 1

Discriminant

= b² - 4ac

= 1² - 4 × 1 × 1

= 1 - 4

= - 3

Step 4 of 4 :

Find number of real roots of the equation

Discriminant

= b² - 4ac

= - 3 < 0

So the equation has no real roots.

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