Question

If the product of the zeroes of the quadratic polynomial kx^2



-9x+6 is 2, then

find the value of k.​

Answer 1

Given :

• Product of zeroes of a polynomial kx² - 9x + 6 as 2

To Find :

• The value of k

Knowledge Required :

• The relation between the coefficients of a quadratic equation and roots of the equation (in terms of x let) is given by ,

$\\ {\boxed{\sf{sum \: of \: the \: roots = \frac{ - (x \: coeffcient)}{( {x}^{2} \: coeffcient) } }}}$

$\\ {\boxed{\sf{product \: of \: the \: roots = \frac{(constant \: term)}{( {x}^{2} \: coeffcient) } }}}$

Solution :

Given quadratic polynomial = kx² - 9x + 6

From the given quadratic polynomial we get ,

• x² coefficient = k
• x coefficient = -9
• constant term = 6

We are given product of zeroes as 2.

Substituting the values we have in the second relation. We get ,

$\\ : \implies \sf \: 2 = \frac{6}{k}$

$\\ : \implies \sf \: 2k = 6$

$\\ : \implies \sf \: k = \frac{6}{2}$

$\\ : \implies \underline{\boxed{\pink{\mathfrak{k = 3}}}} \: \bigstar$

$\\ \therefore{\underline{\sf{Hence \: , \: The \: value \: of \: k \: is \: \bold{3}.}}}$

Answer 2

️️️️️Opekaa️️️️️️!

Solution :

Given quadratic polynomial = kx² - 9x + 6

From the given quadratic polynomial we get ,

x² coefficient = k

x coefficient = -9

constant term = 6

We are given product of zeroes as 2.

Substituting the values we have in the second relation. We get ,

$\huge{\boxed{\bold{Good to know}}}$

product of the roots

=(constant term)/x² coeffcient

$\rm given \longmapsto {x}^{2} 4x + k = 0 \\ \\ \\ \\ \rm \: product \: \: of \: \: zerous = 3 \\ \\ \\ \\ \rm \: we \: \: know \: \: that \\ \\ \\ \\ \rm \tt \sf \bold{product \: \: of \: \: zeroes = \frac{c}{a} = \frac{k}{1} } \\ \\ \\ \\ \longmapsto\boxed{ \tt \:3 = \frac{k}{1} } \\ \\ \\ \\ \longmapsto \boxed{ \blue k = 3}$