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\$

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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