First slide

Theory of equations

Question

If the equation ${ax}^{2} + bx + c = x$ has no real roots, then the equation $a {({ax}^{2} + bx + c)}^{2} + b ({ax}^{2} + bx + c) + c = x$ will be

Easy

A

four real roots
B

no real root
C

at least two real roots
D

None of these

Solution

Since ${ax}^{2} + bx + c = x$ has no real roots
$\Rightarrow {ax}^{2} + bx + c > x or {ax}^{2} + bx + c < x$ for all real x.
Let ${ax}^{2} + bx + c > x and {ax}^{2} + bx + c = P (x)$
$∴ P (x) > x$ for all real x.
$\begin{array}{l} \Rightarrow P (P (x)) > P (x) \\ \Rightarrow a {({ax}^{2} + bx + c)}^{2} + b ({ax}^{2} + bx + c) + c > {ax}^{2} + bx + c > x \end{array}$
Hence, given equation has no real solution.
Similarly it can be proved for ${ax}^{2} + bx + c < x$

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