please answer question number 12 and 14
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take that
m = 3-√3/5
n = 3+√3/5
m+n
= 3-√3/5+3+√3/5
= 3-√3+3+√3/5
= 6/5
m×n
= 3-√3/5×3+√3/5
= 9-3/25
= 6/25
so normal form of quadric equation is
x²-(m+n)x + (mn)
= x²-6x/5+6/25
m = 3-√3/5
n = 3+√3/5
m+n
= 3-√3/5+3+√3/5
= 3-√3+3+√3/5
= 6/5
m×n
= 3-√3/5×3+√3/5
= 9-3/25
= 6/25
so normal form of quadric equation is
x²-(m+n)x + (mn)
= x²-6x/5+6/25
Priya12345678:
Thanks.. :)
14) take that mx²+(m+n)x+n = 0 mx² +mx+nx+n = 0. mx(x+1)+n(x+1) = 0. (x+1) (mx+n) = 0. x+1 = 0. mx+n = 0. x= -1. x= -n/m. so the zeroes of mx²+(m+n)x+n. is. x= -1. or x= -n/m
thanks a lot...
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