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If coefficients of the equation ax2 + bx + c = 0, a ¹ 0 are real and roots of the equation are non-real complex and a + c < b, then
(A) 4a + c > 2b (B) 4a + c < 2b (C) 4a + c = 2b (D) none of these
Please give the solution of this question.
Solution": Let assume x=-1 a-b+c but a+c hence f(x)<0 for all real values of x therefore putting x=-2 we get f(X)=4a+c-2b<0 or 4a+c<2b
Solution": Let assume x=-1
a-b+c but
a+c
hence f(x)<0 for all real values of x
therefore
putting x=-2
we get f(X)=4a+c-2b<0
or 4a+c<2b
nC6:n-3C3=91:4
xy
encoded with the matrix -3 -7 and 4 9. what lights up a soccer stadium? ecoded message: {-3 - 7} {3 2 } {3 6} {57 127} {52 127} {77 173} {23 51)
three towns are situated in such away that town B is 120 kilometers on a bearing of 030 degrees from town A. Town C is 210 kilometers on a bearing of 110 degrees from town A (a)ca
10+2=
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