Quadratic EquationHard
Question
If the equation $ax^{2} + bx + c = x,a,b,c \in R$ and $a \neq 0$, has no real roots, then the equation $a\left( ax^{2} + bx + c \right)^{2} + b\left( ax^{2} + bx + c \right) + c = x$ will have
Options
A.2 distinct real roots
B.no real roots
C.2 equal real roots
D.nothing can be said
Solution
Let $P(x) = {ax}^{2} + bx + c$
$P(x) = x$ has no real roots
$$\begin{matrix} & \ \Rightarrow \ P(x) > x\forall x \in R \\ & \ \Rightarrow \ P(P(x)) > P(x) > x\forall x \in R \end{matrix}\ \text{~or~}\ \begin{matrix} & P(x) < x \\ & P(P(x)) < P(x) < x\forall x \in R \end{matrix}\ \forall x \in R$$
Hence, $P(P(x)) = x$ has no real roots
Create a free account to view solution
View Solution FreeMore Quadratic Equation Questions
The diagram shows the graph of y = ax2 + bx + c. Then -...Let $\alpha,\beta$ be the roots of the equation $x^{2} - 3x + r = 0$, and $\frac{\alpha}{2},2\beta$ be the roots of the ...If roots of x2 − (a − 3)x + a = 0 are such that both of them is greater than 2, then-...If x2 + 2xy + 2x + my − 3 have two rational factors then m is equal to -...Let α, β be the roots of the equation (x - a)(x - b) = c, c ≠ 0 Then the roots of the equation (x - ^...