if $a$ and $b$ are the zero of polynomial $ f(x)=2x^2-9x+9$, then $1/a+1/b$ is equal to :
(a) $9/2$ (b) $3/9$ (c) $1$ (d) $-1$
which is the correct answer?
well actually $a $ and $b$ are alpha and also beta here.
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i resolve upto right here :
$a+b = 9/2$, and also $ab = 9/2$
because alpha + beta = -b/a and alpha*beta = c/a. i tried to reciprocal a+b but that will involved 2/9 yet below the answer is 1 !
and also after that i acquired perplexed.....(answer is: c)
$$frac 1a + frac 1b = frac a+bab$$
That"s sindicate expressing $frac 1a + frac 1b$ as one fractivity after finding the widespread denominator.
That gives you the value $dfracfrac 92frac 92 = 1$, as preferred.
You know that $ab=frac92$ and also $a+b=-frac-92=frac92$. In fact given a 2-degree polynomial $Ax^2+Bx+C$, the sum of the roots is $-fracBA$, while their product is $fracCA$.So$$frac1a+frac1b=fraca+bab=fracfrac92frac92=1$$
HINT$$eginalignfrac1a+frac1b=frac1acdotfrac bb+frac1bcdotfrac aa=fracb+aab.endalign$$
Alternatively: $frac1a$ and also $frac1b$ are the roots of the polynomial $X^2f(frac1X) = 9X^2-9X+2$, so they amount to $1$.
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