utv
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#1
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#1
I'm trying to develop my mathematical understanding

why can't you solve a quadratic such as

x^2+5x+6

by x(X+5)=-6
then
X=-6 and -11 (I know these aren't the real solutions)
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.Matthew.
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#2
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because if you substitute either of those values for X into X(X+5) you don't get -6
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utv
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so this method can only be used for linear equations?
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Muttley79
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#4
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(Original post by utv)
so this method can only be used for linear equations?
What method? Post an example of how you would use this for a linear equation?

To use factorising to solve a quadratic you have to have a zero on one side. Then you know for sure that one of the factors must be zero.
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.Matthew.
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factorising can work for quadratics too, but you need to make the equation equal zero first. for example, the equation you mentioned can be factorised to make (x+2)(x+3)=0.
x can be either -2 or -3, because when you substitute either of these values into the equation, one of the brackets will become zero. and as you know, anything multiplied by zero is also.
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utv
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#6
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oh ok this makes sense as I was wondering why you couldn't factorise sin theta from sin theta squared -2sin theta +1 .
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SerBronn
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Completing the square, and hence the quadratic formula, is essentially doing something similiar but eliminating the x term completely.

x^2 + 5x = (x + {5 \over 2})^2 - ({5 \over 2})^2

Therefore,
x^2 + 5x +6 = 0 \implies (x + {5 \over 2})^2 - ({5 \over 2})^2 + 6 = 0 \implies (x + {5 \over 2})^2 = {25 \over 4} - 6 = {1 \over 4}
Last edited by SerBronn; 2 years ago
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