How do I solve this quadratic square?

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    Quadratic Square
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To solve the quadratic equation ax^2 + bx + c = 0 using the coefficients from the provided numbers (-8, -4, -3, 1, 2, 5, 6, 7, 9), one must ensure that the chosen values yield a valid solution. The initial attempts included selecting coefficients 1, -3, and -4, which worked for one scenario, but other combinations like 1, 2, and 9 did not produce real solutions. A more systematic approach involves recognizing that if a specific number is a solution, it can be used to derive the coefficients through factoring. However, the discussion highlights that factoring is not an option for this problem, complicating the search for valid coefficients. Ultimately, the challenge lies in finding suitable a, b, and c values from the provided set that satisfy the 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?
 
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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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