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Fill in the blanks to make the equation true. ( _x + _) ( _x + _) = _x 2 + _x + _(− 35, − 8, − 5, 1, 3, 4, 6, 7, 12)

 Jun 9, 2024
 #1
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For any a, b, c, d:

\((ax+b)(cx+d) = acx^2 + bcx + adx + bd\)

Now a, c, and ac are all a blank in the equation. This means that a and c may only be certain numbers if the product ac is in the number bank also. The same logic also works with b and d. The only sets of numbers that have this property are (-5, 7) (product is -35) and (3, 4) (product is 12). One of these is (a,c), and the other is (b,d), but we don't know which is which. We can also flip a and b around, to get two more possibilities per equation. This means that there are 4 possibilities, and we can try out each one.

\((-5x + 3)(7x + 4) = -35x^2+(1)x+12\)

\((-5x + 4)(7x + 3) = -35x^2 + 14x + 12\)

\((3x-5)(4x+7) = 12x^2+(1)x-35\)

\((4x-5)(3x+7) = 12x^2 + 14x - 35\)

 

Looking at these equations, the first and third are the only ones that use only numbers in the bank. Therefore, the two solutions are "(-5x+3)(7x+4) = -35x^2 + 1x + 12" and "(3x-5)(4x+7) = 12x^2 + 1x - 35"

 

Please let me know if there is an easier way to do this!

 Jun 9, 2024
edited by Maxematics  Jun 9, 2024
edited by Maxematics  Jun 9, 2024
edited by Maxematics  Jun 9, 2024

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