Question

form a quadratic polynomials p(x) with -1/6 and -1/3 as the sum and product of its zeroes respectively. ​

Answer 1

Solution:-

$\bf{let \: \alpha \: and \: \beta \: be \: the \: zeros \: of \: the \: required \: polynomia \: f(x)}$

then

$( \alpha + \beta ) = \frac{ - 1}{6} \: and \: ( \alpha \beta ) = \frac{ - 1}{3}$

general form of quadratic polynomial is:-

$f(x) = {x}^{2} - ( \alpha + \beta )x + ( \alpha \beta )$

now we have put the value :-

$f(x) = {x}^{2} - ( \frac{ - 1}{6} )x + ( \frac{ - 1}{3} )$

now we have simply the polynomial

$f(x) = {x}^{2} - ( \frac{ - 1}{6} )x + ( \frac{ - 1}{3} )$

take a lcm =6

$f(x) = \frac{6x {}^{2} + 1x + - 1 }{6}$

Quadratic polynomial is:-

$f(x) = \frac{6x {}^{2} + 1x + - 1 }{6}$