# Stuck, Stuck, Stuck - Need Help Urgently

• May 17th 2005, 04:15 AM
Boysey
Stuck, Stuck, Stuck - Need Help Urgently
Hey guys, really struggling today as per usual!!

Can anyone help me with this problem, here goes:

For the quadratic equation 3X2(squared) + 8X - 4 = 0, which two options are the correct solutions rounded to two decimal places.
A) 0.28
B) -0.34
C) 0.43
D) -0.61
E) 2.52
F) -2.79
G) 2.92
H) -3.10

If anyone could help you would really be saving my back.!! Thanks!!
• May 17th 2005, 05:55 AM
hpe
Quote:

Originally Posted by Boysey
Hey guys, really struggling today as per usual!!

Can anyone help me with this problem, here goes:

For the quadratic equation 3X2(squared) + 8X - 4 = 0, which two options are the correct solutions rounded to two decimal places.
A) 0.28
B) -0.34
C) 0.43
D) -0.61
E) 2.52
F) -2.79
G) 2.92
H) -3.10

If anyone could help you would really be saving my back.!! Thanks!!

There are all sorts of ways to do this (just plug the trial solutions into the equation until you have found two correct solutions), but I can't think of any way that doesn't require up to seven attempts to find the first solution and a little more work to find the second one.

If you have a graphing calculator and are allowed to use it, graph the equation and look up the zeroes. This may be a good idea anyway when you have solved the problem with some other method.

Your best bet is to look up the solution formula in your math book and compute the solutions from scratch, then you can find them in the list.
• May 17th 2005, 06:41 AM
ticbol
3x^2 +8x -4 = 0

Are you allowed to use the quadratic formula?
If yes, then,
x = [-8 +,-sqrt(b^2 -4(3)(-4))] / (2*3)
x = [-8 +,-sqrt(112)] / 6
x = (-8 +,-10.58) / 6

x = (-8 +10.58)/6 = 0.43

x = (-8 -10.58)/6 = -3.10