`x ^(2) +1` has no real zeroes. Why ?

If a x^2+b x+c=0,a ,b ,c in R has no real zeros, and if c 0 (c) a >0

To the equation 2^(2pi//cos^(-1)x) - (a + (1)/(2)) 2^(pi/cos^(-1)x) -a^(2) = 0 has only one real root, then

A quadratic trinomial P(x)=a x^2+b x+c is such that the equation P(x)=x has no real roots. Prove that in this case equation P(P(x))=x has no real roots either.

If the equation a x^2+b x+c=x has no real roots, then the equation a(a x^2+b x+c)^2+b(a x^2+b x+c)+c=x will have a. four real roots b. no real root c. at least two least roots d. none of these

If the equation 2x^(4)+x^(3)-2x-1=0 has two real roots alpha and beta , then between alpha and beta the equation 8x^(3)+3x^(2)-2=0 has -

If a, bin{1,2,3} and the equation ax^(2)+bx+1=0 has real roots, then

Let a in R and let f: Rvec be given by f(x)=x^5-5x+a , then (a) f(x) has three real roots if a >4 (b) f(x) has only one real roots if a >4 (c) f(x) has three real roots if a<-4 (d) f(x) has three real roots if -4

Does the polynomial a^4+4a^2+5 have real zeroes ?

Show that the equation e^(x-1)+x-2=0 has no real root greater than 1.

Show that the equation e^(x-1)+x-2=0 has no real root which is less than 1.