here's another i can't solve.

t - 5√t - 14 = 0

i added 14 to both sides and then squared them. then i minused 196 and am left with t^2 + 15t -196 = 0

this doesn't work, what have i done wrong?

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- Dec 9th 2009, 04:26 PMMiasmagasmaEquation with radical
here's another i can't solve.

t - 5√t - 14 = 0

i added 14 to both sides and then squared them. then i minused 196 and am left with t^2 + 15t -196 = 0

this doesn't work, what have i done wrong? - Dec 9th 2009, 04:32 PMskeeter
- Dec 9th 2009, 04:34 PMMiasmagasma
- Dec 9th 2009, 04:55 PMmr fantastic
- Dec 9th 2009, 06:22 PMMiasmagasma
- Dec 9th 2009, 06:31 PMProve It
Notice it looks very much like a quadratic.

So use a dummy variable $\displaystyle x^2 = t$.

This would mean $\displaystyle x = \sqrt{t}$.

Your equation becomes

$\displaystyle x^2 - 5x - 14 = 0$

$\displaystyle (x - 7)(x + 2) = 0$

$\displaystyle x = -2$ or $\displaystyle x = 7$.

Note that, since $\displaystyle \sqrt{t} = x$, the solution $\displaystyle x = -2$ is unusable.

This means, since $\displaystyle t = x^2$, that $\displaystyle t = 49$. - Dec 9th 2009, 06:32 PMmr fantastic
- Dec 9th 2009, 06:48 PMNOX Andrew
As you surmised, $\displaystyle t * 5\sqrt{t} \neq 5t$. On the other hand, $\displaystyle \sqrt{t} * 5\sqrt{t} = 5t$.