Question

One zero of the polynomial 5x²-15x+k is twice the other, then k is​

Answer 1

Solution :

We have quadratic polynomial p(x) = 5x² - 15x + k & zero of the polynomial p(x) = 0.

Let the α & β are the two zeroes of the given polynomial .

$\underline{\underline{\tt{According\:to\:the\:question\::}}}$

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

• a = 5
• b = -15
• c = k

$\mapsto\bf{\alpha =2\beta ....................(1)}$

&

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

$\mapsto\tt{\alpha +\beta =\dfrac{-b}{a} =\bigg\lgroup\dfrac{Coefficient\:of\:x}{Coefficient\:of\:x^{2}} \bigg\rgroup }\\\\\\\mapsto\tt{2\beta + \beta =\dfrac{-(-15)}{5} \:\:\:\:[from(1)]}\\\\\\\mapsto\tt{2\beta + \beta =\cancel{\dfrac{15}{5}} }\\\\\\\mapsto\tt{2\beta + \beta =3}\\\\\\\mapsto\tt{3\beta = 3}\\\\\\\mapsto\tt{\beta = \cancel{3/3}}\\\\\\\mapsto\bf{\beta = 1}$

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

$\mapsto\tt{\alpha =2\times 1}\\\\\mapsto\bf{\alpha= 2}$

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

$\mapsto\tt{\alpha \times \beta =\dfrac{c}{a} =\bigg\lgroup\dfrac{Constant\:term}{Coefficient\:of\:x^{2}} \bigg\rgroup }\\\\\\\mapsto\tt{2 \times 1 =\dfrac{k}{5} }\\\\\\\mapsto\tt{2 =\dfrac{k}{5} }\\\\\\\mapsto\tt{k=5\times 2}\\\\\\\mapsto\bf{k=10}$

Thus;

The value of k will be 10 .