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Question Number 47301 by myear last updated on 08/Nov/18

$$\mathrm{If}\mathrm{the}\mathrm{equation}{x}^2+px+q=0\mathrm{has}\mathrm{roots}$$$$a\mathrm{and}b\mathrm{where}p,q\mathrm{are}\mathrm{non}-\mathrm{zero}$$$$\mathrm{constants}.\mathrm{Then}$$

Answered by Kunal12588 last updated on 08/Nov/18

$$(x-a)(x-b)=x^2+px+q$$$$x^2-(a+b)x+ab=x^2+px+q$$$$comparingbothsides$$$$-(a+b)=p$$$$ab=q$$

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