Question 91699
a)

Remember, real solutions result when the discriminant is greater than or equal to zero. 


So we can use this:


{{{b^2-4ac>=0}}}



note: remember, in the quadratic formula {{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}, the discriminant consists of every term in the square root. Since you cannot take the square root of a negative number, the discriminant must be positive.




Now looking at {{{3x^2+bx-3}}} we see that a=3 and c=-3


{{{b^2-4(3)(-3)>=0}}} Plug in a=3 and c=-3


{{{b^2+36>=0}}} Multiply


{{{b^2>=-36}}} Subtract 36 from both sides



Since {{{b^2}}} is always positive, {{{b^2>=-36}}} is always true. So b can be any real number



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b)

Again, using the discriminant we get:


{{{b^2-4ac>=0}}}


Now looking at {{{5x^2+bx+1}}} we see that a=5 and c=1


{{{b^2-4(5)(1)>=0}}} Plug in a=5 and c=1


{{{b^2-20>=0}}} Multiply


{{{b^2>=20}}} Add 20 to both sides


{{{b>=sqrt(20)}}} or {{{b<=-sqrt(20)}}} Take the square root of both sides


So b can be greater than or equal to {{{sqrt(20)}}} or less than or equal to{{{-sqrt(20)}}} (but not both at the same time)



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Again, using the discriminant we get:


{{{b^2-4ac>=0}}}


Now looking at {{{-3x^2+bx-3}}} we see that a=-3 and c=-3


{{{b^2-4(-3)(-3)>=0}}} Plug in a=-3 and c=-3



{{{b^2-36>=0}}} Multiply


{{{b^2>=36}}} Subtract 36 from both sides


{{{b>=6}}} or {{{b<=-6}}} Take the square root of both sides


So b can be greater than 6 or less than -6 (but not both at the same time)



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d)

The general rule (the one we've been using) is the discriminant {{{b^2-4ac}}} must be greater than or equal to zero in order to produce real solutions. Remember, you cannot take the square root of a negative number, so the discriminant must be positive. 



So the general rule is 



{{{b^2-4ac>=0}}}