How do I solve this quadratic square?

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SUMMARY

The discussion centers on solving the quadratic equation ax² + bx + c = 0 using specific coefficients selected from the numbers -8, -4, -3, 1, 2, 5, 6, 7, and 9. The user initially attempted to find coefficients through trial and error, successfully identifying a combination of 1, -3, and -4 for the first row, yielding a valid solution. However, the user struggled with the first column's coefficients of 1, 2, and 9, which resulted in complex solutions. A more efficient strategy suggested involves recognizing that if a solution exists, it can be expressed as a factor of the quadratic equation.

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Homework Statement


Basically I have to fill out this ax^2+bx+c using the given numbers
-8,-4,-3,1,2,5,6,7,9
The solution must be the one provided

So it works like this (example is just with regular addition)

1 2 3=5
8
9
=
18

Here is the problem:

http://www.shareapic.net/content.php?id=24056338&owner=sheldon3

Edit: pick the coefficients a, b, and c from the nine numbers at the top of the page (-8, -4, -3, 1, 2, 5, 6, 7, and 9) so that the equation ax2 + bx + c = 0 has the number to the right or below as a solution.

Homework Equations



ax^2+bx+c

x=-b+- √(b)^2 - 4(a)(c) / 2 QUADRATIC FORMULA

The Attempt at a Solution



I found the top row to be 1,-3,-4
And the first column to be 1,2,9

I accomplished this using trial and error. Is there a more efficient way to solve this?
 
Last edited:
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You haven't explained this very well, IMO, and the picture you posted doesn't, either. What I think is going on (which you should confirm) is that you're supposed to pick the coefficients a, b, and c from the nine numbers at the top of the page (-8, -4, -3, 1, 2, 5, 6, 7, and 9) so that the equation ax2 + bx + c = 0 has the number to the right or below as a solution.

This works for the first row, for which you picked a, b, and c to be 1, -3, and -4. The equation x2 - 3x =4 = 0 has x = -1 for a solution.

You're choices for a, b, and c of 1, 2, and 9 in the first column don't work, if we're talking about the equation x2 + 2x + 9 = 0, since -.2280 is not a solution of this equation. In fact the only solutions of this equation are complex, and these aren't possible choices.

A strategy that's not as much trial and error is to notice that, for example, if 2.3508 is a solution of the quadratic, then x - 2.3508 is a factor. This means that you have (x - 2.3508)(ax + d) = 0, or ax2 + (d - 2.3508)x - 2.3508d = 0.

At this point, you want to find a number d so that b = (d - 2.3508) and c = -2.3508d are among the 9 numbers that you're given to work with.
 
Yea your explanation is right. Thanks
Anyway the work states that you cannot solve by factoring.
I kind of understand what you're saying. But if I need c= -2.3508d, that wouldn't be a number which I was given.
 

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