Discussion between harry and Yiyuan Lee

Stack Exchange

16:07

A: complex numbers for quadratic equation

If $a, b$ are roots of a quadratic equation, then the quadratic equation can be written as $$(x-a)(x-b) = 0$$ This is intuitively true because substituting in either of the roots $a,b$ will cause LHS to yield $0$. If you want a more rigorous proof, this result is known as the Weierstrass facto...

harry

but how do you know (x-a)(x-b)=0?

Yiyuan Lee

Substitute in $a$ into LHS. Then it becomes $(a-a)(a-b) = 0\cdot(a-b) = 0$.

harry

Thanks~ so root can be (x-a)(x-b)~

wait, but why (x+a) (x+b) = 0 ?

Yiyuan Lee

If you took $(x+a)(x+b) = 0$, then the roots would be $x=-a$ and $x=-b$ instead. This is because if you substitute (either of ) these two values in, you will observe that the LHS really resolves into $0$.

harry

but it will become x^2 - 5x + 7 + i = 0

Yiyuan Lee

16:07

Then that will be the quadratic equation you were looking for. :)

harry

really? i thought the answer is x^2 + (-5+i)x + 8 - i = 0 (but maybe the answer sheet is wrong)

Yiyuan Lee

You might want to try expanding it out again. Hold on, I'll improve my answer to show you a shortcut.

I believe the answer sheet is incorrect. WolframAlpha verifies : wolframalpha.com/input/?i=x%5E2+-+5x+%2B+7+%2B+i

harry

thanks very much :D