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Given eq x^2-6x+3=0  what is (u^2+u-12)/(u^2+5u+6)+(t^2+t-12)/(t^2+5t+6)? u,t roots of...

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greenbel | Honors

Posted July 7, 2013 at 4:24 PM via web

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Given eq x^2-6x+3=0 

what is (u^2+u-12)/(u^2+5u+6)+(t^2+t-12)/(t^2+5t+6)?

u,t roots of equation

don't calculate the roots

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted July 7, 2013 at 4:40 PM (Answer #1)

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You need to replace the roots u and t in the given equation, such that:

`u^2 - 6u + 3 = 0 => u^2 = 6u - 3`

`t^2 - 6t + 3 = 0 => t^2 = 6t - 3`

Replacing `6u - 3` and `6t - 3` for `u^2` and `t^2` , in the relation that you need to evaluate it, yields:

`(u^2 + u - 12)/(u^2 + 5u + 6) + (t^2 + t - 12)/(t^2 + 5t + 6) = (6u - 3 + u - 12)/(6u - 3 + 5u + 6) + (6t - 3 + t - 12)/(6t - 3 + 5t + 6)`

`(u^2 + u - 12)/(u^2 + 5u + 6) + (t^2 + t - 12)/(t^2 + 5t + 6) = (7u - 15)/(11u + 3) + (7t - 15)/(11t + 3) `

`(u^2 + u - 12)/(u^2 + 5u + 6) + (t^2 + t - 12)/(t^2 + 5t + 6) = ((7u - 15)(11t + 3) + (7t - 15)(11u + 3))/((11u + 3)(11t + 3))`

`(u^2 + u - 12)/(u^2 + 5u + 6) + (t^2 + t - 12)/(t^2 + 5t + 6) = (77ut + 21u - 165t - 45 + 77ut + 21t - 45 - 165u)/(121ut + 33(u+t) + 9)`

`(u^2 + u - 12)/(u^2 + 5u + 6) + (t^2 + t - 12)/(t^2 + 5t + 6) = (154ut + 21(u+t) - 165(u+t) - 90)/(121ut + 33(u+t) + 9)`

`(u^2 + u - 12)/(u^2 + 5u + 6) + (t^2 + t - 12)/(t^2 + 5t + 6) = (154ut - 144(u+t) - 90)/(121ut + 33(u+t) + 9)`

You need to use Vieta's formulas, such that:

`u + t = -(-6)/1 => u + v = 6`

`ut = 3/1 = 3`

Replacing `6` for `u+t` and `3` for `ut` yields:

`(u^2 + u - 12)/(u^2 + 5u + 6) + (t^2 + t - 12)/(t^2 + 5t + 6) = (154*3 - 144*6 - 90)/(121*3 + 33*6 + 9)`

`(u^2 + u - 12)/(u^2 + 5u + 6) + (t^2 + t - 12)/(t^2 + 5t + 6) = (462 - 864 - 90)/(363 + 198 + 9)`

`(u^2 + u - 12)/(u^2 + 5u + 6) + (t^2 + t - 12)/(t^2 + 5t + 6) = -492/570 = -82/95`

Hence, evaluating the given expression, using Vieta's formulas, yields `(u^2 + u - 12)/(u^2 + 5u + 6) + (t^2 + t - 12)/(t^2 + 5t + 6) = -82/95.`

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