Question

find the zeroes of the qudratic polynomia :6x²-x-2
​

Answer 1

Step-by-step explanation:

• 6x² - x - 2 = 0
• 6x² + 4x - 3x - 2 = 0
• 2x ( 3x + 2 ) - 1 ( 3x - 2 ) = 0
• ( 2x - 1 ) + ( 3x + 2 ) = 0

Two zeroes :-

$\red{~▼△▼△▼△▼△▼△▼△▼△▼△▼△▼△▼△▼~}$

→ 2x - 1 = 0

→ 2x = 1

→ x = ½

$\blue{~▼△▼△▼△▼△▼△▼△▼△▼△▼△▼△▼△▼△~}$

→ 3x + 2 = 0

→ 3x = -2

→ x = $\frac{-2}{3}$

$\red{~▼△▼△▼△▼△▼△▼△▼△▼△▼△▼△▼△▼△~}$

$\alpha$ = ½ and $\beta$ = $\frac{-2}{3}$

We know that

$\frac{ \alpha }{ \beta } + \frac{ \beta }{ \alpha }$

$\frac{ \frac{1}{2} }{ \frac{ - 2}{3} } + \frac{ \frac{ - 2}{3} }{ \frac{1}{2} }$

$= > ( \frac{1}{2} \times \frac{3}{ - 2} ) \: + \: ( \: \frac{ - 2}{3} \times \frac{2}{1} )$

$= > \frac{ - 3}{4} \: + \: \frac{4}{3}$

$= > \frac{ - 9 - 16}{12} \\ \\ = > \frac{ - 25}{12}$

Thus ,

$\alpha$ + $\beta$ = $\frac{-25}{12}$