Question

If one root of the quadratic equation 2x^2+kx-6=0 is 2 then find the value of k also find the other root

Answer 1

Given, p(x) = 2x² + kx -6
and p(2) = 0

=> p(2) = 2(2)² + k(2) - 6 = 0
=> 8 + 2k - 6 = 0
=> 2 + 2k = 0
=> k = -1
Now, dividing p(x) by ( x - 2 ), we get p(x) = (x-2)(2x+3)
Hence, other root is 2x+3 = 0 => x = (-3/2)

Hope, u r satisfies....... Plz mark brainliest.....

Answer 2

$\huge \boxed{ \underline{ \underline{ \bf{Answer}}}}$

Since x = 2 is a root of the given equation

2x² + kx - 6= 0

2 × 2² + 2k - 6 = 0

8 + 2k - 6 = 0

2k + 2 = 0

k = -1

Now put k = -1 in the equation 2x² + kx - 6= 0, than we get :-

2x² + kx - 6= 0

⇒ 2x² - 4x + 3x - 6 = 0

⇒ 2x ( x-2 ) + 3 ( x-2 ) = 0

⇒ ( x - 2 ) ( 2x + 3 ) = 0

⇒ x - 2 = 0    , 2x + 3 = 0

⇒ x = 2 , x = -3/2

Hence, the other root is -3/2.