how to find $t$ from $2t^2-0.01t^4=100$?

Question

how to find $t$, from $2t^2-0.01t^4=100$? I was guessing may be I can take $t^2$ common but if it is so so why cannot we take $t$ common in other cases? I mean, for example: $t^2+4t=-4$ why can we find the answer like $t(t+4)=-4$ therefore the answer is $t=-4$

Answer 1

Please see the EDIT below: For $2t^4-0.01t^4=100$: Write $(2-.01)t^4=1.99t^4=100$. Then, divide by $1.99$ on both sides and take a fourth root. Do not forget to write $\pm$, as $4$ is an even power.

You cannot solve your example as you have shown. You must apply the zero product property, which says if $ab=0$, then $a$, $b$, or both are $0$. Instead, write $t^2+4t+4=(t+2)^2=0$.

EDIT (see comment below): For $2t^2-.01t^4=100 \Rightarrow 2t^2-.01t^4-100=0$. Then, let $t^2=x$. This change in variable gives $2x-.01x^2-100=-.01x^2+2x-100=0$. This is a quadratic equation. Can you solve it?

Answer 2

$.01t^4-2t^2+100=0$ $\rightarrow(.1t^2-10)^2=0$ $\rightarrow t^2=100$ $\rightarrow t=10\ or\ t=-10$

Answer 3

$t^2=u$ where $2 u -0.01 u^2=100$. This is a quadratic equation in $u$.

Answer 4

You seem to be asking if we have $x(x + b) = c$ why is that we can sometimes say $x = c$ and why sometimes we can't and when is it the right way to do it and when is it the wrong way to do it.

The reason we can sometimes do it is because we sometimes get lucky and pull the right answer out of our ass. Sometimes we can't do it because sometimes we aren't lucky and we don't pull the right answer out of our ass. When is it the right way to do it? Never. And when is it the wrong way to do it? Always.

If $x(x+b) = c$ then there are two factors of $c$ (call them $c_1$ and $c_2$) and $x = c_1$ and also $x+b = c_2$. The problem is, we have no @#!&ing way of knowing which of the infinite pairs of possible $c_1$s and $c_2$s will work. Don't get me wrong. There will be one pair which work. But there will be an infinite number that do not work.

So we could do $x(x + 4) = -4 = -4*1$ and figure $x = -4$ and $x + 4 =1$. The only thing wrong with doing that is it's dead wrong. Or we could do $x(x + 4) = -4 = -2*2$ and figure $x = -2$ and $x + 4 = 2$. The only thing right about this is that it happens to be the right answer. Or we could do $x(x+4) = 593*-4/593$ and figure $x = 593$ and $x + 4 = -4/593$. That's also a wrong answer.

So how do we know which one to do? We don't!

So why did $t^2(2 - 0.01t^2) = 100 = 100*1$ with $t^2 = 100$ and $2 - 0.01t^2 = 1$ work? Because we were lucky. Sometimes you sit in a boat and a fish just jumps in your lap.

BUT there are two times this will actually work.

If $x(x + b) = 0$. Whereas $c \ne 0$ will have an infinite number of factor pairs. $c = 0$ has only one; $0$ and $0$.

So if $x(x+b) =0$ we know there are two possible solutions $x = 0$ or $x = -b$.

The other possible situation that will work is $x(x + 0) = c$. Then we know $x = \pm \sqrt{c}$.

So the right way to do it is to force one of those.

In other words, complete the square or use the quadratic formula:

$t^2 + 4t = -4 \implies t^2 + 4t + 4 = 0 \implies (t + 2)^2 = 0 \implies t = -2$

Or

$2t^2 - 0.01t^4 = 100 \implies .01t^4 - 2t^2 + 100 = 0 \implies (.1t^2 - 10)^2 = 0 \implies t^2 = 100 \implies t = \pm 10$.