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31 Aug 2009, 06:01
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31 Aug 2009, 06:11
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its c--->
y=0=x^2+x( a+b)+ab from stmt 1
-6 = ab from 2 stmt
so y = x^2 - x -6 = 0
thus calculate the x,y values
y=0=x^2+x( a+b)+ab from stmt 1
-6 = ab from 2 stmt
so y = x^2 - x -6 = 0
thus calculate the x,y values
Updated on: 31 Aug 2009, 09:15
Originally posted by CasperMonday on 31 Aug 2009, 06:52.
Last edited by CasperMonday on 31 Aug 2009, 09:15, edited 1 time in total.
Last edited by CasperMonday on 31 Aug 2009, 09:15, edited 1 time in total.
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I solved it:
\(y=(x+a)(x+b)=x^2+ax+bx+ab\)
That statement (1) is insufficient, is clear.
Now, if we plug in the coordinates given in statement (2), we get
\(-6=0+0+0+ab\)
Combined with statement (2), we get a system of two equations,
\(a+b=1\)
\(ab=-6\)
Solve it and get the answer. Hence, C.
\(y=(x+a)(x+b)=x^2+ax+bx+ab\)
That statement (1) is insufficient, is clear.
Now, if we plug in the coordinates given in statement (2), we get
\(-6=0+0+0+ab\)
Combined with statement (2), we get a system of two equations,
\(a+b=1\)
\(ab=-6\)
Solve it and get the answer. Hence, C.
