Plz. solve as soon as possible Q-5
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if x=1/2√3 is zero of p(x)
=>2k(1/2√3)^2-7(1/2√3)+k=0
2k(1/12)-7/2√3+k=0
k/6-7/2√3+k=0
12√3k-42+12√3k/12√3=0
24√3k-42=0
24√3k=42
k=42/24√3
k=7/4√3
=>2k(1/2√3)^2-7(1/2√3)+k=0
2k(1/12)-7/2√3+k=0
k/6-7/2√3+k=0
12√3k-42+12√3k/12√3=0
24√3k-42=0
24√3k=42
k=42/24√3
k=7/4√3
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