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For what value of b are the roots of 3x^2 + bx - 5 = 0 complex.
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The roots of a quadratic equation ax^2 + bx + c = 0 are `(-b+-sqrt(b^2-4ac))/(2a)` . These are complex if b^2 - 4ac < 0
For the equation 3x^2 + bx - 5 = 0 the roots are complex if:
b^2 - 4*-5*3 < 0
=> b^2 + 60 < 0
=> b^2 < -60
But the square of a number is never negative.
For no value of b are the roots of the equation 3x^2 + bx - 5 = 0 complex.
Posted by justaguide on April 9, 2013 at 4:26 AM (Answer #1)
3x^2 + bx - 5 = 0
The roots of the equation `3x^2+bx-5=0` are
``Let us suppose x is complex no.
So x will complex if
which is contradiction to fact that b is real no .Also there does not exist order relation in complex no. system . So our assumption is wrong.
So this equation has no complex root for any real value of b.
Posted by pramodpandey on April 9, 2013 at 4:37 AM (Answer #2)
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