Find value of p for which one root of quadratic eqn px²-14x+8=0 is 6 times than the other
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Answer:
Step-by-step explanation:
px^2-14x+8=0
let a and b be the roots of the equation
b=6a
sum of roots =-b/a
product of roots = c/a
here a=p , b=-14 , c=8
a+b=14/p
ab=8/p
a+6a=14/p
7a=14/p
a=2/p
a*6a=8/p
6a^2=8/p
3a^2=4/p
3*(2/p)^2=4/p
3*4/p^2=4/p
p=3
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