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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
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The digraph D for a relation R on V = {1, 2, 3, 4} is shown below (a) show that (V,R) is a poset. (b) Draw its Hasse diagram. (c) Give a total order that have R.
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