How many real solutions does the equation b2 + 8b − 3 = 0 have?
A. 0
B. 1
C. 2
D. 3
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Nature of roots can be found using determinant (D)
D=b^2-4ac
Where a=1, b=8, c=-3
D=8^2-(1)*(-3)
=64+3
=67
Since D>0, number of roots is 2.
If D=0, number of roots is 1 (both roots are equal)
If D<0, number of roots is 0 (no real roots)
D=b^2-4ac
Where a=1, b=8, c=-3
D=8^2-(1)*(-3)
=64+3
=67
Since D>0, number of roots is 2.
If D=0, number of roots is 1 (both roots are equal)
If D<0, number of roots is 0 (no real roots)
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