Find two numbers that add up to 8 and multiply to 20. Only the complex number "i" (imaginary number) is allowed other than the real numbers. But you do not necessarily have to use "i" if it is unnecessary. Just putting out the possibilities
I've been doing trial and error for ages and I can't find any. Tried decimals, fractions, etc... What is a good way to go about solving this?
$\endgroup$4 Answers
$\begingroup$Let $x$ and $y$ be the two numbers we want. Then $x + y = 8$ and $xy = 20$. Using the first equation, we have that $y = 8-x$. Plugging this into the second equation gives $x(8 - x) = -x^2 + 8x = 20$. Then we have $-x^2 + 8x - 20 = 0$. Using the quadratic formula we get $x = 4 \pm 2i$. Then $y = 4\mp 2i$. Specifically, $x$ and $y$ should be complex conjugates of each other.
$\endgroup$ 0 $\begingroup$Hint :
$(a-b)^2=(a+b)^2-4ab$
where $a+b=8$ and $ab=20$
$\endgroup$ $\begingroup$Hint $\ a,b, $ are roots of $\,(x\!-\!a)(x\!-\!b) = x^2 -(a\!+\!b) x + ab.\,$ But $\, a+b = 8\ $ and $\, ab = 20\ $ so $\ldots$
$\endgroup$ $\begingroup$$$(a-b)^2 = a^2 - 2 ab + b^2 = a^2 + 2 ab + b^2 -4 ab = (a+b)^2-4ab = 64-4*20 = -16$$
So $$a-b=4 i$$
Add the two $$ a+b=8\\a-b=4i$$ to get $$2a=8+4i \text { or } a = 4+2i$$ Subtract to get $$2b = 8-49 \text { or } b = 4-2i$$
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