Breaking large radicals (Intermediate Algebra)

[Drifter]

How do I Simplify a large radical like this
( I couldn't find the radical symbol so I used a dash)

/5905 or /9816

Original problem 0= 4.9t^2+4t-500
This is the problem I am trying to solve written in quadratic formula

( x= -b(+-) / b-4ac // 2a )

This is the problem set in Quadratic Formula
-4(+/-)/9816
9.8

[Rahul@gurome]

Solution:
Given equation is 4.9t^2+4t-500 = 0.
Or 49t^2 + 40t - 5000 = 0.

Roots are [-b+ sqrt{(b^2)-4*a*c}]/(2*a) and [-b - sqrt{(b^2)-4*a*c}]/(2*a).

Replace the values with a = 49, b = 40 and c = -5000,

Now b^2-4*a*c = 1600 +4*49*5000 = 400(4 + 49*5*10) = 400(2454).
So sqrt{(b^2)-4*a*c} = 20*sqrt(2454).

So roots are {-40 + 20*sqrt(2454)}/(2*49) and {-40 - 20*sqrt(2454)}/(2*49)
or {-20 + 10*sqrt(2454)}/49 and {-20 - 10*sqrt(2454)}/49.

Since 2454 is not a perfect square, you can't get rational numbers as answers.

[selango]

Drifter,

Can you pls post the original problem?

[Ian Stewart]

How do I Simplify a large radical like this
( I couldn't find the radical symbol so I used a dash)

/5905 or /9816

Original problem 0= 4.9t^2+4t-500
This is the problem I am trying to solve written in quadratic formula

( x= -b(+-) / b-4ac // 2a )

This is the problem set in Quadratic Formula
-4(+/-)/9816
9.8

I'd just add a few points:

* You never, ever, need to use the Quadratic Formula on the GMAT. Of course, you sometimes could use it, but there will always be faster approaches. If you think you need the Quadratic Formula for a GMAT question, you've missed a trick somewhere.

* You also don't need to simplify square roots of crazy numbers. There's just no fast way to do that, so they can't ask you to. You will never need to simplify the square root of 9816 (as it turns out, you can only factor out a 4 anyway).

* You do, however, need to simplify square roots of simple numbers, by dividing out perfect squares. So if you see sqrt(50), you need to rewrite that as sqrt(25)*sqrt(2) = 5*sqrt(2).

* For larger numbers, you can always figure out how to simplify a square root by finding the prime factorization of the number under the root. If we need to simplify sqrt(180), one way we can do this it to prime factorize 180 to get sqrt(2^2 * 3^2 * 5). We can then see all the perfect squares we can take out: sqrt(180) = sqrt(2^2)*sqrt(3^2)*sqrt(5) = 6*sqrt(5). There are of course other ways to do this as well.

[selango]

Agree with Ian..

Drifter Can you please post the original question..surely there ll be some other approach to solve this problem.