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.

You are watching: The sum of the values of α and β

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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