how do you solve 3u^2-4=0 with no decimal approximations

Question

amistre64 · Answer

by not making it a decimal

anonymous · Answer

but how do you solve it?

amistre64 · Answer

show me the point where you get stuck at ...

amistre64 · Answer

or walk me thru it to the point your stuck at

anonymous · Answer

I don't even know where to start

amistre64 · Answer

maybe adding something to both sides would help start it off ....

amistre64 · Answer

or, consider this option ...

(what?), minus 4, equals 0 ?

anonymous · Answer

so would you change it to $$3u^2=4$$

amistre64 · Answer

yes;  now consider:

3 times (what?) = 4

anonymous · Answer

well, wouldn't it have to be a mixed number?

amistre64 · Answer

yes; but lets keep it as an improper fraction

amistre64 · Answer

try dividing both sides by some value ... to get rid of that 3

anonymous · Answer

ok, so $$u^2=4/3$$

anonymous · Answer

so you would take the square root of each side?

amistre64 · Answer

very good, now at this point is where people tend to start having troubles ... yes!!  but there are 2 results for it:  a + and a -

amistre64 · Answer

$$u=\pm\sqrt{4/3}$$

amistre64 · Answer

we can still simplify this more

anonymous · Answer

yes, the final answer must be in exact form

amistre64 · Answer

most times the do not like to have a sqrt on the bottom of a fraction:$$u=\pm\sqrt{\frac43}$$
$$u=\pm\frac{\sqrt4}{\sqrt3}$$
$$u=\pm\frac{\sqrt4}{\sqrt3}\frac{\sqrt3}{\sqrt3}$$
$$u=\pm\frac{\sqrt4\sqrt3}{\sqrt3^2}$$
can you see where its going from there?

anonymous · Answer

can't you cancel the 3?

amistre64 · Answer

you can cancel the sqrt^2 and leave the 3

amistre64 · Answer

$$u=\pm\frac23 \sqrt3$$

anonymous · Answer

ohhhh ok!