Math, asked by jomilojuajagbe, 1 year ago

Given that one of the roots of the equation 2x^2+(k+2)x+k=0 is 2. Find the value of k

Answers

Answered by Pratishtha2003

putting x = 2

2 (2)² + ( k+2) 2 + k = 0

2(4) + 2k + 4 + k = 0

8 + 3k + 4 = 0

3k = -12

k = -4

hope this helps if so mark as brainliest.

Answered by lalitc2502

Answer:

-4

Step-by-step explanation:

since 2 is one of the roots,

⇒2x²+(k+2)x+k=0

⇒2(2)²+(k+2)(2)+k=0

⇒8+2k+4+k=0

⇒12+3k=0

⇒3k= -12 (by taking from lhs to rhs or by subtracting 12 from both sides)

⇒k = -12/3 = -4

jomilojuajagbe: Can it be solved like a quadratic equation. Thanks

lalitc2502: yes you can, just do it as 2 (since it is the root) = (-b± root (b^2-4ac)) /

lalitc2502: 2a

lalitc2502: sorry the 2a was supposed to be with the first comment but i pressed enter by accident

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