Question

If a and b are zero of a quadratic polynomial x²-6x+1 then value of (1/a+1/b-ab ) is

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{\frac{1}{a} + \frac{1}{b} - ab = 5 }}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given: }} \\ \tt: \implies {x}^{2} - 6x +1 = 0 \\\\ \tt:\implies a\:and\:b\:are\:zeroes \\\\ \red{\underline \bold{To \: Find: }} \\ \tt: \implies \frac{1}{a}+\frac{1}{b}-ab= ?$

• According to given question :

$\tt: \implies {x}^{2} - 6x + 1 = 0 \\ \\ \tt \circ \: a = 1 \\ \\ \tt \circ \: b = - 6 \\ \\ \tt \circ \: c = 1 \\ \\ \bold{For \: finding \: value} \\ \tt: \implies \frac{1}{a} + \frac{1}{b} - ab \\ \\ \tt: \implies \frac{b + a -( ab)^{2} }{ab} \\ \\ \tt: \implies \frac{a + b - {(ab)}^{2} }{ab} - - - - - (1) \\ \\ \bold{For \: sum \: of \: zeroes} \\ \tt: \implies sum \: of \: zeroes = \frac{ - b}{a} \\ \\ \tt: \implies a + b = \frac{ - ( - 6)}{1} \\ \\ \green{\tt: \implies a + b= 6} \\ \\ \bold{For \: product \: of \: zeroes} \\ \tt: \implies product \: of \: zeroes = \frac{c}{a} \\ \\ \tt: \implies ab = \frac{1}{1} \\ \\ \green{\tt: \implies ab = 1}\\ \\ \text{Putting \: given \: values \: in \: (1)} \\ \tt: \implies \frac{a + b - {(ab)}^{2} }{ab} \\ \\ \tt: \implies \frac{6 - {1}^{2} }{1} \\ \\ \tt: \implies \frac{6 - 1}{1} \\ \\ \green{\tt: \implies 5} \\ \\ \green{\tt \therefore \frac{1}{a} + \frac{1}{b} - ab = 5}$

Answer 2

$\begin{lgathered}\green{\underline \bold{Given: }} \\ \tt: \implies {x}^{2} - 6x +1 = 0 \\\\ \tt:\implies a\:and\:b\:are\:zeroes \\\\ \red{\underline \bold{To \: Find: }} \\ \tt: \implies \frac{1}{a}+\frac{1}{b}-ab= ?\end{lgathered}$

Given:

• According to given question :

$</p><p>\begin{lgathered}\tt: \implies {x}^{2} - 6x + 1 = 0 \\ \\ \tt \circ \: a = 1 \\ \\ \tt \circ \: b = - 6 \\ \\ \tt \circ \: c = 1 \\ \\ \bold{For \: finding \: value} \\ \tt: \implies \frac{1}{a} + \frac{1}{b} - ab \\ \\ \tt: \implies \frac{b + a -( ab)^{2} }{ab} \\ \\ \tt: \implies \frac{a + b - {(ab)}^{2} }{ab} - - - - - (1) \\ \\ \bold{For \: sum \: of \: zeroes} \\ \tt: \implies sum \: of \: zeroes = \frac{ - b}{a} \\ \\ \tt: \implies a + b = \frac{ - ( - 6)}{1} \\ \\ \green{\tt: \implies a + b= 6} \\ \\ \bold{For \: product \: of \: zeroes} \\ \tt: \implies product \: of \: zeroes = \frac{c}{a} \\ \\ \tt: \implies ab = \frac{1}{1} \\ \\ \green{\tt: \implies ab = 1}\\ \\ \text{Putting \: given \: values \: in \: (1)} \\ \tt: \implies \frac{a + b - {(ab)}^{2} }{ab} \\ \\ \tt: \implies \frac{6 - {1}^{2} }{1} \\ \\ \tt: \implies \frac{6 - 1}{1} \\ \\ \green{\tt: \implies 5} \\ \\ \green{\tt \therefore \frac{1}{a} + \frac{1}{b} - ab = 5}\end{lgathered}$