Question

Quadratic polynomial 6x^2-7x+2 has zeros as alpha and beta now form a quadratic polynomial whose zeros are 5alpha and 5 beta

Answer 1

Solution :

We have quadratic polynomial p(x) = 6x² - 7x + 2 & zero of the polynomial p(x) = 0

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

• a = 6
• b = -7
• c = 2

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

$\mapsto\sf{\alpha +\beta = \dfrac{-b}{a} =\bigg\lgroup\dfrac{Coefficient\:of\:x}{Coefficient\:of\:x^{2}}\bigg \rgroup }\\\\\\\mapsto\sf{\alpha + \beta =\dfrac{-(-7)}{6} }\\\\\\\mapsto\bf{\alpha + \beta =\dfrac{7}{6}}$

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

$\mapsto\sf{\alpha \times \beta = \dfrac{c}{a} =\bigg\lgroup\dfrac{Constant\:term}{Coefficient\:of\:x^{2}}\bigg \rgroup }\\\\\\\mapsto\sf{\alpha \times \beta =\dfrac{2}{6} }\\\\\\\mapsto\bf{\alpha \times \beta =\dfrac{2}{6}}$

$\underbrace{\sf{According\:to\:the\:question\::}}}$

We have two zeroes 5α & 5β, so;

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

$\mapsto\sf{\alpha +\beta = \dfrac{-b}{a} =\bigg\lgroup\dfrac{Coefficient\:of\:x}{Coefficient\:of\:x^{2}}\bigg \rgroup }\\\\\\\mapsto\sf{5\alpha + 5\beta }\\\\\\\mapsto\sf{5(\alpha + \beta )}\\\\\\\mapsto\sf{5\times \dfrac{7}{6} }\\\\\\\mapsto\bf{\dfrac{35}{6} }$

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

$\mapsto\sf{\alpha \times \beta = \dfrac{c}{a} =\bigg\lgroup\dfrac{Constant\:term}{Coefficient\:of\:x^{2}}\bigg \rgroup }\\\\\\\mapsto\sf{5\alpha \times 5\beta }\\\\\\\mapsto\sf{25(\alpha \times \beta )}\\\\\\\mapsto\sf{25\times \dfrac{2}{6} }\\\\\\\mapsto\sf{\cancel{\dfrac{50}{6} }}\\\\\mapsto\bf{25/3}}$

Thus;

The new quadratic equation required;

$\longrightarrow\sf{x^{2} - (sum\:of\:zeroes)x + (product\:of\:zeroes)}\\\\\longrightarrow\sf{x^{2} - \bigg(\dfrac{35}{6} \bigg)x + \bigg(\dfrac{25}{3} \bigg)}\\\\\longrightarrow\bf{6x^{2} - 35x + 50}$

Answer 2

Step-by-step explanation:

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