Suppose a quadratic equation whose roots are 3 and -2,
so the equation is (x-3)(x+2) = 0
So, from here we get,
x-3 = 0
and,
x+2 = 0
Since, both equations equal to zero, we can equate these two,
x-3 = x+2
So, -3 = 2 ?
Where did I go wrong?
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3You used AND instead of OR. – hamam_Abdallah Nov 03 '16 at 18:42
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I don't get it. Why shouldn't it be 'AND'? – Sachin Chaudhary Nov 03 '16 at 18:48
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$0*3=0$ but $3\neq 0$ – hamam_Abdallah Nov 03 '16 at 18:50
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I don't get you. – Sachin Chaudhary Nov 03 '16 at 18:51
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I have derived x - 3 = x + 2 , why is this equation wrong? – Sachin Chaudhary Nov 03 '16 at 18:51
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$x-3=0$ IF AND ONLY IF $x=3$. – hamam_Abdallah Nov 03 '16 at 18:54
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now you got it? – hamam_Abdallah Nov 03 '16 at 22:52
1 Answers
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You are wrong when you say that
x-3=0 AND x+2=0
You correctly reduced the given quadratic to
(x-3)(x+2)=0
Hence this equation will be satisfied if and only if any ONE of the conditions are satisfied....ie....Either
x-3=0
OR
x+2=0
(Since 0x(any no.)=0)
Hence you get two solutions
x=3
AND
x=-2
Obviously x can take only a single value at a time.
SirXYZ
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Thank you so much. I understood it. And can you explain that why do we write the quadratic equation as (x-a)(x-b)=0 if a and b are the two roots. I have just learned it. I don't know the proof/logic – Sachin Chaudhary Nov 03 '16 at 19:08
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We just factorise the given second degree expression into two factors.....from which we easily get the solutions of the given quadratic. Obviously all such expressions cannot be factorised into rational/integral factors....but anyways thats another case....which you will learn in coming years I guess. – SirXYZ Nov 03 '16 at 19:12