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20 Feb 2006, 14:32
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Does G(x)=ax^2+bx+c have 2 intersections with axis-x?
1) G(x)-4 has 2 intersections with axis-x
2) G(x-4) has 2 intersections with axis-x
1) G(x)-4 has 2 intersections with axis-x
2) G(x-4) has 2 intersections with axis-x
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20 Feb 2006, 17:45
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The answer is B .
Before i explain both the case the imp rule to remember is that in order for equation G(x) to have 2 intersection b^2 -4ac >0
case 1 G(x)-4 = ax^2+bx+ (c-4) has 2 intersction
that means b^2 -4 *a *(c-4) > 0
means b^2 -4ac +16a >0
since we dont know anything about the value of a itsdifficult to make out whether b^2 -4ac will be > 0 so the data is insufficient .
case 2 g(x-4) = a * ((x-4)^2) + b(x-4) +c
=a* x^2 -x (8a-b) +16a-4b+c
from the above equation
A= a
B= b-8a
C =16a-4b+c
since case 2 has 2 intersection means B^2 -4AC > 0
==> (b-8a) ^2 - 4 *a*(16a-4b+c) > 0
==> b^2 -16ab+64a^2 -64a^2 +16ab-4ac >0
==>b^2 -4ac >0
it means the equation G(x) will have two intersection with x axis
Before i explain both the case the imp rule to remember is that in order for equation G(x) to have 2 intersection b^2 -4ac >0
case 1 G(x)-4 = ax^2+bx+ (c-4) has 2 intersction
that means b^2 -4 *a *(c-4) > 0
means b^2 -4ac +16a >0
since we dont know anything about the value of a itsdifficult to make out whether b^2 -4ac will be > 0 so the data is insufficient .
case 2 g(x-4) = a * ((x-4)^2) + b(x-4) +c
=a* x^2 -x (8a-b) +16a-4b+c
from the above equation
A= a
B= b-8a
C =16a-4b+c
since case 2 has 2 intersection means B^2 -4AC > 0
==> (b-8a) ^2 - 4 *a*(16a-4b+c) > 0
==> b^2 -16ab+64a^2 -64a^2 +16ab-4ac >0
==>b^2 -4ac >0
it means the equation G(x) will have two intersection with x axis
sorry i am a bit dumb... pls explain dear 
I jus solved it using the -b +- sqrt(b^2-4a.c)/ 2a formula. Thanks a lot dear. l.o.k.s
