SOLUTION: Find the two values of "b" for which the quadratic equation 4x^2+bx+49=0 has two real solutions b=( ) b= ( )

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Question 617410: Find the two values of "b" for which the quadratic equation 4x^2+bx+49=0 has two real solutions
b=( ) b= ( )
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
A quadratic equation will have two real solutions if its discriminant, b%5E2-4ac, is positive. IOW:
b%5E2-4ac+%3E+0
Inserting our values for a and c:
b%5E2-4%284%29%2849%29+%3E+0
b%5E2-784+%3E+0
b%5E2+%3E+784
Since 28%5E2+=+784, b can be any number greater than 28 or less than -28! As long as b > 28 or b < -28, there will be two real solutions to your equation. Just pick any two such numbers.

There not just two possible values for b as the problem seems to suggest.