Question

3x^2 +kx+2=0
please help it's urgent​

Answer 1

root, it must satisfy the equation.

substitute x=2 in the given equation,

we get 3(2)^2-k(2)-2=0

12–2k-2=0

2k=10

so, k=5

Now substitute k value in the equation, we get 3x^2–5x-2=0

By factorization we can write it as 3x^2–6x+x-2=0

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

(x-2)(3x+1)=0

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

x=2, x=-1/3

so other root is -1/3

ANSWER: k=5 and the other root is -1/3

SOLUTION:

3x^2-kx-2 = 0 ……….,.Eq.(A)

Since ‘2’ is one of the roots of the Eq.(A), we can write:

3(2)^2-k(2)-2 =0

=> 12–2k-2 = 0

=> 10–2k =0

=> 2k = 10

=> k =5

By substituting the value of ‘k’ in Eq.(A), we get:

3x^2–5x-2 = 0

=> 3x^2–6x+1x-2 = 0

=> 3x(x-2)+ 1(x-2) = 0

=> (3x+1)(x-2) =0

=> 3x+1 =0 (or) x-2 =0

=> 3x= -1 (or) x =2

=> x= -1/3 (or) x =2

Therefore, the other root is -1/3.

Put that root value in equation and find the value of k,after getting k put it in original equation and factorize it

3x^2-kx-2=0………………..(1)

Root is x=2

Putting x in equation 1

3(4)-2k=2

12–2k=2

-2k=-10

k=5

Now put k in equation 1

3x^2–5x-2=0

(3x+1)(x+2)=0

x=2 andx=-1/3 and value of k is 5

If the sum of the root of the equation 3x2+kx+1=0 is 7, then what is the