If `P(x) = x^2 + ax + b` and `Q(x) = x^2 + a_1x + b_1, a, b, a_1, b_1, in R` and equation `P(x).Q(x)` has at most one real root, then

If P(x)=x^2+ax+b and Q(x)=x^2+a_1x+b_1,a,b,a_1,b_1 epsilon R and equation P(x).Q(x)=0 has at most one real root, then (A) (1+a+b)(1+a_1+b_1)gt0 (B) (1+a+b)(1+a_1+b_1)lt0 (C) (1+a+b)/(1+a_1_b_1)gt0 (D) 1+a+bgt0

If P(x)=x^2+ax+b and Q(x)=x^2+a_1x+b_1,a,b,a_1,b_1 epsilon R and equation P(x).Q(x)=0 has at most one real root, then (A) (1+a+b)(1+a_1+b_1)gt0 (B) (1+a+b)(1+a_1+b_1)lt0 (C) (1+a+b)/(1+a_1_b_1)gt0 (D) 1+a+bgt0

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

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.

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.

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.

Let P(x) =x^(2)+ 1/2x + b and Q(x) = x^(2) + cx + d be two polynomials with real coefficients such that P(x) Q(x) = Q(P(x)) for all real x. Find all the real roots of P(Q(x)) =0.

If a_1x^2 + b_1 x + c_1 = 0 and a_2x^2 + b_2 x + c_2 = 0 has a common root, then the common root is

Let P={1,2,3} , Q={a, b} Find P x Q and Q x P