Question

Roots of quadratic equation 6x^2-x-2=0 are *

2/3, 1/2

2/3, - 1/2

-2/3, - 1/2

​

Answer 1

Answer:

-1/2,2/3

Step-by-step explanation:

6x

2

−x−2=0

6x

2

−4x+3x−2=0

2x(3x−2)+(3x−2)1=0

(2x+1)(3x−2)=0

2x+1 = 0 or 3x-2 = 0

x = -1/2 , x=2/3.

Answer 2

AnswEr:-

❀ Your Answer Is Option 2) 2/3, -1/2.

ExplanaTion:-

Given quadratic equation:-

$\\ \bf \hookrightarrow 6 {x}^{2} - x - 2 = 0.$

To Find:-

• The roots of the given quadratic equation.

So Now,

Lets find out the roots by factorization method:-

Here,

$\\ \rm\leadsto coeffiicient \: \: of \: \: {x}^{2} \times constant \: \: term. \\ \\ \tt = 6 \times 2. \\ \\ \bf = 12. \\ \\ \rm and \\ \\ \bf \leadsto12 = 2 \times 2 \times3. \\ \\ \rm also \\ \\ \bf\leadsto(2 \times 2) - 3 = 1 = \rm middle \: \: term.$

So Lets split the middle term:-

$\\ \tt\mapsto6 {x}^{2} - x - 2 = 0. \\ \\ \tt\mapsto6 {x}^{2} - (4 - 3)x - 2 = 0. \\ \\ \tt\mapsto6 {x}^{2} - 4x + 3x - 2 = 0. \\ \\ \rm(by \: \: op ening \: \: brackets). \\ \\ \tt\mapsto2x(3x - 2) + 1(3x - 2) = 0. \\ \\ \tt\mapsto(3x - 2)(2x + 1) = 0. \\ \\ \tt\mapsto3x - 2 = 0 \: \: \: \: or \: \: \: \: 2x + 1 = 0. \\ \\ \tt\mapsto 3x = 2 \: \: \: \: or \: \: \: \: 2x = - 1. \\ \\ \bf \hookrightarrow \boxed{ \bf x = \dfrac{2}{3} \: \: \: \: or \: \: \: \: x = - \dfrac{1}{2}.}$

❝ $\bf\therefore$ The roots of the given equation are $\bf\dfrac{2}{3},\:\:-\dfrac{1}{2}.$❞

So The Final Answer Is Option 2.