sir this is quadratic equations based on common root problems
Asked by Aravindh Nandha Raj s | 30th May, 2013, 07:01: PM
Expert Answer:
Let m be the common root. Then,
m^2+am+12=0,
m^2+bm+15=0,
a = -(m^2+12)/m
b = -(m^2+15)/m
Also, m^2+(a+b)m+15=0
Substituting the value of a and b
m^2+(
-(m^2+12)/m +
-(m^2+15)/m
)m+15=0
m^2 -2m^2-27+15 = 0
m^2 = -13
m = sqrt(13)i
a = -(m^2+12)/m = 1/sqrt(13)i = -sqrt(13)i/13
b = -(m^2+15)/m = -2/sqrt(13)i = -2sqrt(13)i/13
m^2+am+12=0,
m^2+bm+15=0,
a = -(m^2+12)/m
b = -(m^2+15)/m
Also, m^2+(a+b)m+15=0
Substituting the value of a and b
-(m^2+12)/m +
-(m^2+15)/m
)m+15=0a = -(m^2+12)/m = 1/sqrt(13)i = -sqrt(13)i/13
b = -(m^2+15)/m = -2/sqrt(13)i = -2sqrt(13)i/13
Answered by | 31st May, 2013, 04:55: AM
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