Question

6x2+x-2 find zeroes and verify ​

Answer 1

Answer:

- $\frac{2}{3}$ and 0.5

Step-by-step explanation:

f(x) = ax² + bx + c = a( x - $x_{1}$ )( x - $x_{2}$ )

$x_{1}$ , $x_{2}$ are zeroes of f(x).

$x_{1}$ = $\frac{-b+\sqrt{b^2 -4ac} }{2a}$

$x_{2}$ = $\frac{-b-\sqrt{b^2 -4ac} }{2a}$

D = b² - 4ac

~~~~~~~~~~~~~~~

f(x) = 6x² + x - 2

D = 1 - 4( 6 )( - 2 ) = 49 = 7²

$x_{1}$ = ( - 1 + 7 ) ÷ 12 = 0.5

$x_{2}$ = ( - 1 - 7 ) ÷ 12 = - $\frac{2}{3}$

Verification:

6( x - 0.5 )( x + $\frac{2}{3}$ ) = ( x - 0.5 )( 6x + 4 ) = 6x² - 3x + 4x - 2 = 6x² + x - 2