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Posted December 30th, 2009 by alpha

Find the least value of $\frac{6x^2-22x+21}{5x^2-18x+17 }$ for all real values of x, using the theory of quadratic equations.

Comments

On January 2nd, 2010 Structure says:

$5x^2-18x+17>0$
if for all $x\in R$

$\frac{6x^2-22x+21}{5x^2-18x+17}\ge a$

$(6-5a)x^2+2(9a-11)x+21-17 a\ge 0$

You must have complex roots and $6-5a>0$
so $\Delta\le 0$

$(9a-11)^2-(6-5a)(21-17a)\le 0$

or $4a^2-9a+5\ge 0$
$(4a-5)(a-1)\ge 0$
$a\in (-\infty ,1]\cup [\frac{5}{4},+\infty )$ $
so as $a< \frac{6}{5}$
you have greatest a=1