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Let f(x) = ax2+bx+c, where a,b,c belongs to R and a is not equals to zero. If f(x) = x has non real roots, then show that the eqn f(f(x)) = x has all non real roots.

Tushar Watts , 15 Years ago

Grade 12

3 Answers

Jitender Singh

Last Activity: 10 Years ago

Ans:
Hello Student,
Please find answer to your question below

$f(x) = ax^2 + bx + c$
$f(x) =x$
$ax^2 + bx + c = x$
$ax^2 + (b-1)x + c = 0$
This equation have non real roots.
(b – 1)²< 4ac
$f(f(x)) = x$
$a(f(x))^2 + 2bf(x) + c = x$
$a(ax^2 + bx + c)^2 + 2b(ax^2 + bx + c) + c = x$
$a(ax^4 + b^2x^2 + c^2 + 2abx^3 + 2bcx + 2acx^2) + 2b(ax^2 + bx + c) + c = x$ $a^2x^4 + (b^2 + 2ac + 2ab)x^2 + (c^2 + 2bc + c) + 2abx^3 + (2bc-2b ^2-1)x = 0$
$a^2x^4 + 2abx^3 + (b^2 + 2ac + 2ab)x^2 + (2bc-2b^2-1)x + (c^2 + 2bc + c) = 0$
So this equation also have non real roots.

mycroft holmes

Last Activity: 10 Years ago

Without Loss of Generality we can assume that a>0.

Then f(x) = x has no real roots implies that f(x)> x for all real x.

This also implies that f(f(x))>x for all real x.

This means f(f(x)) – x = 0 has non-real roots.

mycroft holmes

Last Activity: 10 Years ago

I need to add an explanation:

f(x)> x for all x implies f(f(x))>f(x)>x. That’s how we get the inequality f(f(x))>x for all x

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