• Dec 9th 2009, 04:26 PM
Miasmagasma
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 PM
skeeter
Quote:

Originally Posted by Miasmagasma
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?

$\displaystyle t-14 = 5\sqrt{t}$

$\displaystyle t^2 - 28t + 196 = 25t$

$\displaystyle t^2 - 53t + 196 = 0$
• Dec 9th 2009, 04:34 PM
Miasmagasma
Quote:

Originally Posted by skeeter
$\displaystyle t-14 = 5\sqrt{t}$

$\displaystyle t^2 - 28t + 196 = 25t$

$\displaystyle t^2 - 53t + 196 = 0$

cheers but why did my method not work?
• Dec 9th 2009, 04:55 PM
mr fantastic
Quote:

Originally Posted by Miasmagasma
cheers but why did my method not work?

Because $\displaystyle (t - 5 \sqrt{t})^2 \neq t^2 + 15t$. If you post all your working on how you got that (wrong) result your error(s) can be explained.
• Dec 9th 2009, 06:22 PM
Miasmagasma
Quote:

Originally Posted by mr fantastic
Because $\displaystyle (t - 5 \sqrt{t})^2 \neq t^2 + 15t$. If you post all your working on how you got that (wrong) result your error(s) can be explained.

well i thought t times 5√t = 5t which is no doubt wrong. how do you work that out?
• Dec 9th 2009, 06:31 PM
Prove It
Quote:

Originally Posted by Miasmagasma
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?

Notice it looks very much like a quadratic.

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

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

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

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

$\displaystyle x = -2$ 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 PM
mr fantastic
Quote:

Originally Posted by Miasmagasma
well i thought t times 5√t = 5t which is no doubt wrong. how do you work that out?

How can $\displaystyle t \cdot 5 \sqrt{t} = 5t$? If t = 4 do you get a correct result from this? And what then do you think $\displaystyle \sqrt{t} \cdot 5 \sqrt{t}$ is equal to?
• Dec 9th 2009, 06:48 PM
NOX Andrew
As you surmised, $\displaystyle t * 5\sqrt{t} \neq 5t$. On the other hand, $\displaystyle \sqrt{t} * 5\sqrt{t} = 5t$.