Question

HELP ME GUYS YOU ARE GREAT.... ​

Answer 1

ANSWER

$\boxed{ \large{ {x}^{2} - \frac{ \large{7}} { \large{2}}x + 3 } }$

STEP BY STEP EXPLANATION

Given,

6y² - 7y + 2

6y² - 3y - 4y + 2

3y(2y - 1) -2(2y - 1)

(3y - 2)(2y - 1)

3y - 2 = 0

3y = 2

$\large{y = \frac{ \large{2}}{ \large{3}} }$

2y - 1 = 0

2y = 1

$\large{y = \frac{ \large{1}}{ \large{2}} }$

$\alpha = \large{\frac{ \large{2}}{ \large{3}} }$

$\beta = \large{ \frac{ \large{1}}{ \large{2}} }$

$\large{ \frac{ \large{1}}{ \large{ \alpha }} = \large{\frac{ \large{3}}{ \large{2}} }}$

$\large{ \frac{ \large{1}}{ \large{ \beta }} = \large{{ \large{2}} }}$

Equation whose zeroes will

$\large{ \frac{ \large{1}}{ \large{ \alpha }} } \: and \: \frac{ \large{1}}{ \large{ \beta }} \: are$

$\large{ {x}^{2} - ( \frac{1}{ \alpha } + \frac{1}{ \beta } )x + \frac{1}{ \alpha \beta } }$

$\large{ {x}^{2} - ( \frac{3}{ 2} + {2} )x + \frac{3 \times 2}{ 2 } }$

${ \large{ {x}^{2} - \frac{ \large{7}} { \large{2}}x + 3 } }$

So,

Equation whose zeroes will

$\large{ \frac{ \large{1}}{ \large{ \alpha }} } \: and \: \frac{ \large{1}}{ \large{ \beta }} \: are$

$\boxed{ \large{ {x}^{2} - \frac{ \large{7}} { \large{2}}x + 3 } }$