Question 11.
$(x^{2} + 1)^{2} - x^{2} = 0$ has:

(a) four real roots

(b) two real roots

(c) no real roots

(d) one real roots

Open in App

Solution

Option (C)
Given, $(x^{2} + 1)^{2} - x^{2} = 0$
$\Rightarrow (x^{2} + 1)^{2} = x^{2}$
Square root on both sides. We get,
$x^{2} + 1 = x o r x^{2} + 1 = - x$
$\Rightarrow x^{2} - x + 1 = 0 o r x^{2} + x + 1 = 0$
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.

Suggest Corrections

Similar questions

Question 8
The quadratic equation $2 x^{2} - \sqrt{5} x + 1 = 0$ has
(A) two distinct real roots
(B) two equal real roots
(C) no real roots
(D) more than 2 real roots

The nature of roots of $x^{2} - 3 x + 2 = 0$ will be:

Q. The equation

e^{s i n x} - e^{- s i n x} - 4 = 0

has:

Q. If

a, b, c, d \in R,

then the equation

(x^{2} + a x - 3 b) (x^{2} - c x + b) (x^{2} - d x + 2 b) = 0

has

Q. The equation

e^{x - 1}

+ x - 2 = 0 has

Join BYJU'S Learning Program

Explore more

Solving a Quadratic Equation by Completion of Squares Method

Standard X Mathematics

Join BYJU'S Learning Program

Question 11. (x2+1)2−x2 = 0 has: (a) four real roots (b) two real roots (c) no real roots (d) one real roots

Option (C) Given, (x2+1)2−x2 = 0 ⇒(x2+1)2=x2 Square root on both sides. We get, x2+1=x or x2+1=−x ⇒x2−x+1=0 or x2+x+1=0 discriminant for the above equations will be; 1-4 = -3 <0 or 1-4=-3 <0 Hence, (x2+1)2−x2 = 0 has no real roots.

Question 11.
$(x^{2} + 1)^{2} - x^{2} = 0$ has:

(a) four real roots

(b) two real roots

(c) no real roots

(d) one real roots