Question

show that (x^2+1)^2 - x^2 = 0 has no real roots.

Answer 1

(x^2+1)^2-x^2=0
(x^2+1)^2=x^2
rooting on both sides
we get
 x^2+1=x       x^2+1=-x
x^2-x+1=0      x^2+x+1=0
det of equations is less than zero
hence they have imaginary roots

Answer 2

Answer: Showed.

Step-by-step explanation: We are given to show that the following equation has no real roots.

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

We know that, for a quadratic equation $ax^2+bx+c=0(a\neq 0)$, the discriminant 'D' is defined as

$D=b^2-4ac.$

The classification of the roots of the equation, based on the discriminant 'D' is as follows:

(i) The roots are real and equal id D = 0.

(ii) The roots are real and unequal if D > 0.

(iii) The roots are unreal if D < 0.

Let, x² = y, then our given equation becomes

$(y+1)^2-y=0\\\\\Rightarrow y^2+y+1=0.$

Here, a = 1, b = 1  and  c = 1.

Therefore, the discriminant is given by

$D=b^2-4ac=1^2-4\times 1\times 1=1-4=-3<0.$

Thus, the given equation has no real roots.