if x=3by2 is the zeros of the polynomial 2xSq+kx-12 then find the value of k.
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Solution :
Given x = 3/2
Let p(x) = 2x² + kx -12
if x = 3/2 is a zero of p(x) then,
p(3/2) = 0
=> 2(3/2)² + k(3/2) - 12 = 0
=> 9/2 + 3k/2 - 12 = 0
=> ( 3k + 9 )/2 = 12
=> 3k + 9 = 24
=> 3k = 24 - 9
=> 3k = 15
=> k = 15/3
=> k = 5
••••••
Given x = 3/2
Let p(x) = 2x² + kx -12
if x = 3/2 is a zero of p(x) then,
p(3/2) = 0
=> 2(3/2)² + k(3/2) - 12 = 0
=> 9/2 + 3k/2 - 12 = 0
=> ( 3k + 9 )/2 = 12
=> 3k + 9 = 24
=> 3k = 24 - 9
=> 3k = 15
=> k = 15/3
=> k = 5
••••••
Answered by
0
Answer:
Step-by-step explanation:
Given,
2x²+kx-12=0
2(3/2)²+(3/2)k-12=0
(9/2)+(3/2)k-12=0
(3/2)k=12-(9/2)
(3/2)k=15/2
∴ k=5
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