Given that one of the roots of the equation 2x^2+(k+2)x+k=0 is 2. Find the value of k
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Answered by
1
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.
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
1
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
yes you can, just do it as 2 (since it is the root) = (-b± root (b^2-4ac)) /
2a
sorry the 2a was supposed to be with the first comment but i pressed enter by accident
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