# Math Help - hard fraction

1. That's better, although there's no real need to reduce to decimal unless you're asked to do so. Technically, you should write

$x_{1}=\dfrac{1+\sqrt{57}}{2}\approx 4.27,$ and

$x_{2}=\dfrac{1-\sqrt{57}}{2}\approx -3.27.$

You don't have equality there, but a decimal approximation. Make sense?

I would reduce to decimal if it is difficult to interpret the final answer in terms of a decimal. But you have a rough order of magnitude here that's straightforward: $7<\sqrt{57}<8,$ so you can see where the answers are.

2. Originally Posted by Ackbeet
That's better, although there's no real need to reduce to decimal unless you're asked to do so. Technically, you should write

$x_{1}=\dfrac{1+\sqrt{57}}{2}\approx 4.27,$ and

$x_{2}=\dfrac{1-\sqrt{57}}{2}\approx -3.27.$

You don't have equality there, but a decimal approximation. Make sense?

I would reduce to decimal if it is difficult to interpret the final answer in terms of a decimal. But you have a rough order of magnitude here that's straightforward: $7<\sqrt{57}<8,$ so you can see where the answers are.
we are just asked to pu it like this:
$x_{1}=\dfrac{1+\sqrt{57}}{2}\approx 4.27,$ and

$x_{2}=\dfrac{1-\sqrt{57}}{2}\approx -3.27.$

a very BIG thankyou for the help in this topic!!!!!
would you like me to that all the the posts you made which helped?

3. You're very welcome.

As for thanking posts which helped, I think it adds value to the forum if you thank just those posts which are really helpful. If something is not helpful, then don't thank it. That's my philosophy. That way, when guests peruse the forum, they can see what sorts of things were helpful and what sorts of things weren't.

It's completely up to you, though. Have a good one!

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