polynomials

**may** · June 5th 2010, 05:50 PM

if the root of the equation 3x^2-6x+1=0 are a and b,find the value of

[a^2+1/(b^2)][b^2+1/(a^2)]

**Prove It** · June 5th 2010, 06:06 PM

Originally Posted by may

if the root of the equation 3x^2-6x+1=0 are a and b,find the value of

[a^2+1/(b^2)][b^2+1/(a^2)]

First of all, I can't tell if this is

$\left(\frac{a^2 + 1}{b^2}\right)\left(\frac{b^2 + 1}{a^2}\right)$ or $\left(a^2 + \frac{1}{b^2}\right)\left(b^2 + \frac{1}{a^2}\right)$ .

Anyway, to begin to answer your questions...

$3x^2 - 6x + 1 = 0$

$x = \frac{6 \pm \sqrt{(-6)^2 - 4(3)(1)}}{2(3)}$

$= \frac{6 \pm \sqrt{36 - 12}}{6}$

$= \frac{6 \pm \sqrt{24}}{6}$

$= \frac{6 \pm 2\sqrt{6}}{6}$

$= \frac{3 \pm \sqrt{6}}{3}$ .

So $a = \frac{3 - \sqrt{6}}{3}$ and $b = \frac{3 + \sqrt{6}}{3}$ . Substitute these values into your expression and simplify.

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