SOLUTION: decide all values of b in the following equations that will give one or more real number solutions. a. 3x^2+bx-3=0 b. 5x^2+bx+1=0 c. -3x^2+bx-3=0 d. Write a rule for judging

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: decide all values of b in the following equations that will give one or more real number solutions. a. 3x^2+bx-3=0 b. 5x^2+bx+1=0 c. -3x^2+bx-3=0 d. Write a rule for judging      Log On


   

Question 91699: decide all values of b in the following equations that will give one or more real number solutions.

a. 3x^2+bx-3=0
b. 5x^2+bx+1=0
c. -3x^2+bx-3=0
d. Write a rule for judging if an equation has solutions by looking at it in standard form.
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
a)
Remember, real solutions result when the discriminant is greater than or equal to zero.

So we can use this:

b%5E2-4ac%3E=0


note: remember, in the quadratic formula x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+, 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%5E2%2Bbx-3 we see that a=3 and c=-3

b%5E2-4%283%29%28-3%29%3E=0 Plug in a=3 and c=-3

b%5E2%2B36%3E=0 Multiply

b%5E2%3E=-36 Subtract 36 from both sides


Since b%5E2 is always positive, b%5E2%3E=-36 is always true. So b can be any real number


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b)
Again, using the discriminant we get:

b%5E2-4ac%3E=0

Now looking at 5x%5E2%2Bbx%2B1 we see that a=5 and c=1

b%5E2-4%285%29%281%29%3E=0 Plug in a=5 and c=1

b%5E2-20%3E=0 Multiply

b%5E2%3E=20 Add 20 to both sides

b%3E=sqrt%2820%29 or b%3C=-sqrt%2820%29 Take the square root of both sides

So b can be greater than or equal to sqrt%2820%29 or less than or equal to-sqrt%2820%29 (but not both at the same time)


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Again, using the discriminant we get:

b%5E2-4ac%3E=0

Now looking at -3x%5E2%2Bbx-3 we see that a=-3 and c=-3

b%5E2-4%28-3%29%28-3%29%3E=0 Plug in a=-3 and c=-3


b%5E2-36%3E=0 Multiply

b%5E2%3E=36 Subtract 36 from both sides

b%3E=6 or b%3C=-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%5E2-4ac 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%5E2-4ac%3E=0