Question

If the sum of the zeroes of the quadratic polynomial 3x2

-kx+6 is 3, then find

the value of k.​

Answer 1

S O L U T I O N :

We have quadratic polynomial p(x) = 3x² - kx + 6 & zero of the polynomial, p(x) = 0

As we know that α & β are the two zeroes of the given polynomials.

A/q

α = 3

As we know that given quadratic polynomial compared with ax² + bx + c;

• a = 3
• b = -k
• c = 6

$\underline{\mathcal{SUM\:OF\:THE\:ZEROES\::}}$

$\mapsto\tt{\alpha +\beta = \dfrac{-b}{a} =\bigg\lgroup \dfrac{Coefficient\:of\:x}{Coefficient\:of\:x^{2}} \bigg\rgroup }$

$\mapsto\tt{\alpha +\beta = \dfrac{-(-k)}{3}}$

$\mapsto\tt{\alpha +\beta = \dfrac{k}{3}..............(1)}$

$\underline{\mathcal{PRODUCT\:OF\:THE\:ZEROES\::}}$

$\mapsto\tt{\alpha \times \beta = \dfrac{c}{a} =\bigg\lgroup \dfrac{Constant\:term}{Coefficient\:of\:x^{2}} \bigg\rgroup }$

$\mapsto\tt{3 \times \beta = \cancel{\dfrac{6}{3}}}$

$\mapsto\tt{3 \times \beta = 2}$

$\mapsto\bf{\beta = 2/3}$

∴ Putting the value of β in equation (1),we get;

$\mapsto\tt{\alpha +\beta = \dfrac{k}{3} }$

$\mapsto\tt{3+\dfrac{2}{3} = \dfrac{k}{3} }$

$\mapsto\tt{\dfrac{9+2}{3} = \dfrac{k}{3} }$

$\mapsto\tt{\dfrac{11}{3} = \dfrac{k}{3} }$

$\mapsto\tt{3k = 33\:\:\underbrace{\sf{cross-multiplication}}}$

$\mapsto\tt{k = \cancel{33/3}}$

$\mapsto\bf{k = 11}$

Thus,

The value of k will be 11 .