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

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      Log On


   



Question 102279: 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

Found 2 solutions by stanbon, MathLover1:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
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
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Each of these is a quadratic equation.
To have Real Number solutions the discriminant must be > or = to zero.
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a) a=3,b=b,c=-3
b^2-4ac>=0
b^2-4*3*-3>=0
b^2+36 >=0
b^2 >= -36
But b^2 is always >= -36 so b may have any Real Number value.
======================
Follow the same procedure on questions b) and c).
Cheers,
Stan H.

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
In the standard form
ax%5E2+%2Bbx+%2Bc+=+0++
the roots are:x=-b+%2B-+sqrt+%28b%5E2+-4%2Aa%2Ac+%29%29+%2F+%282%2Aa%29
if
%28b%5E2-4ac%29%3E=+0
then the roots are real. Otherwise, the roots are imaginary
So
(a) %28b%5E2+-+4%2A3%2A3%29%3E=+0
+b%5E2+%3E=36
+b%3E=+6+ the roots are real for any vale of b+%3E=+6
(b) %28b%5E2+-+4%2A5%2A1%29%3E=+0
b%5E2+%3E=%284%2A5%29
b+%3E=+sqrt%284%2A5%29
b+%3E=+2%2A+sqrt%285%29 the roots are realfor any vale of b+%3E=+2%2A+sqrt+%285%29
(c) %28b%5E2+-+%284%2A-3%2A-3%29%29%3E=+0
b%5E2+%3E=+36
b+%3E=+sqrt+%2836%29
b+%3E=++6 the roots are real for any value of b+%3E=+6