iHateHegartyMath
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Find two integers whose sum is 4 and product is -45
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mqb2766
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(Original post by iHateHegartyMath)
Find two integers whose sum is 4 and product is -45
Look at the factors of 45 and its fairly straightforward.
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kerry0013
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-5×9
9-5
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iHateHegartyMath
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(Original post by kerry0013)
-5×9
9-5
Cheers Kerry!
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kerry0013
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(Original post by iHateHegartyMath)
Cheers Kerry!
um have I say something wrong? is that comment sarcastic!? I mean I'm just replying a question...
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Muttley79
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Please read the rules and do not post solutions
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iHateHegartyMath
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(Original post by kerry0013)
um have I say something wrong? is that comment sarcastic!? I mean I'm just replying a question...
Im not being sarcastic, just saying thank you
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RogerOxon
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(Original post by iHateHegartyMath)
Find two integers whose sum is 4 and product is -45
xy=-45

x+y=4

\therefore x+y=x+\frac{-45}{x}=4

Rearrange to a quadratic, and solve.
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kerry0013
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(Original post by iHateHegartyMath)
Im not being sarcastic, just saying thank you
oh great then... it's the least I can do❤❤❤ lots of love
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Sir Cumference
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(Original post by RogerOxon)
xy=-45

x+y=4

\therefore x+y=x+\frac{-45}{x}=4

Rearrange to a quadratic, and solve.
But to solve the quadratic you need to find two numbers which add to 4 and multiply to -45 It's quicker to just look for the numbers.
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Muttley79
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(Original post by RogerOxon)
xy=-45

x+y=4

\therefore x+y=x+\frac{-45}{x}=4

Rearrange to a quadratic, and solve.
Algebraic approach is not needed here ,,,
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RogerOxon
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(Original post by Sir Cumference)
But to solve the quadratic you need to find two numbers which add to 4 and multiply to -45 It's quicker to just look for the numbers.
True, but a quadratic can be solved by using the quadratic formula. It's a general approach that will work with any numbers - they just happen to be easy here.
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Muttley79
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(Original post by RogerOxon)
True, but a quadratic can be solved by using the quadratic formula. It's a general approach that will work with any numbers - they just happen to be easy here.
This is from a Foundation paper so that approach isn't needed; the formula is really for questions with non-integer solutions.
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RogerOxon
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(Original post by Muttley79)
This is from a Foundation paper so that approach isn't needed; the formula is really for questions with non-integer solutions.
I agree, although people sometimes panic in exams, or miss the obvious, so it's good to know the algebraic approach.
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