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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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IN THIS WE HAVE TO ADD THE PROBABILITY of 3 and 5 occuring separtely and subtract prob. of 3 and 5 occuring together therefore p=(166+100-33)/500=233/500=0.466
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32/562
compare: 643,251: 633,512: 633,893. The answer is 633,512.
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