Question

the roots of the quadratic equation 2 x square minus x minus 6 equal to zero are​

Answer 1

Answer :-

Roots of the equation 2x² - x - 6 = 0 are 2 and - 3/2.

Explanation :-

Given equation

2x² - x - 6 = 0

Comparing 2x² - x - 6 = 0 with 2x² - x - 6 = 0 we get

• a = 2

• b = - 1

• c = - 6

Discriminant (D) = b² - 4ac = (-1)² - 4(2)(-6) = 1 + 48 = 49

D > 0

So the equation has two distinct real roots

Finding roots of the equation using formula

x = (-b ± √D)/2a

⇒ x = {- (-1) ± √49}/2(2)

⇒ x = (1 ± 7)/4

⇒ x = (1 + 7)/4 or x = (1 - 7)/4

⇒ x = 8/4 or x = - 6/4

⇒ x = 2 or x = - 3/2

∴ the roots of the equation 2x² - x - 6 = 0 are 2 and - 3/2.

Answer 2

Given

quadratic equation is

$2 {x}^{2} - x - 6 = 0$

To find

Roots of the given quadratic equation

Explanation

let us calculate the Discriminant

$\underline{\sf D=b^2-4ac}$

in the given quadratic equation

a=2

b=-1

c=-6

$\sf D=(-1)^2-4(2)(-6)$

$\sf D=1+48=49>0$

$\sf so \: roots \: are \: real \: and \: distinct$

We can use the factorisation method directly

$2 {x}^{2} - x - 6 = 0$

$= > 2{x}^{2} - 4x + 3x - 6 = 0$

$= > 2x(x - 2) + 3(x - 2) = 0$

$= > (x - 2)(2x + 3) = 0$

$= > x - 2 = 0 \\ \boxed{ \sf \: x = 2}$

$= > 2x + 3 = 0 \\ \boxed{ \sf \: x = \dfrac{ - 3}{2} }$

Some more information:

If the roots are not real that means

Discriminant is less than 0 then

$\sf x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$