Messages posted by sboosy

How can the product of the 3 distinct integers be a prime number ..because it disobeys the basic rule that prime number has just 2 factors ..that is 1 and itself

Assume that there were 100 persons in all ...so that we can forget the percentages for a while

now 70 people lost their eyes,75 lost their arms,80 lost their ears,85 lost their legs

lets start with the biggest number ....those who lost their legs

now ,

1. 85 people lost their legs

2.now out of the 80 who lost their ears , let 15 of them be the ones who hadnt lost their legs ....

which means minimum 65 lost both their legs and ears

3.now out of the 75 who lost their arms, let 35 of them be the ones who hadnt lost both(legs and ears)

which means that atleast 40 of them lost their legs,ears and arms

4.now out of the 70 who lost their eyes,let 60 of them be the ones who hadnt lost three of their limbs(legs ears

arms) which means that atleast 10 of them lost all 4 limbs

Thus the conclusion is that 10 of them lost all 4 limbs

Question 3:

$ax^2+bx+c=0 \ \mbox{No real roots} \ \Rightarrow D<0 \\ \\ \Rightarrow b^2<4ac \\ \\ \mbox{Given} \ a+c<b \ \Rightarrow (a+c)^2<b^2<4ac \\ \\ \Rightarrow a^2+c^2<2ac \\ \\ \mbox{which is an absurdity as we know} \\ \\ (a-c)^2\geq0 \ \Rightarrow a^2+c^2\geq 2ac$

Question 1:

$x^2-2mx+m^2-1=0 \\ \\ \Rightarrow (x-m)^2-1=0 \ \Rightarrow (x-m-1)(x-m+1)=0 \\ \\ \Rightarrow x=m+1,m-1 \\ \\ \mbox{Now} \ m+1<4 \ \mbox{and} \ m-1>-2 \\ \\ \Rightarrow m\epsilon (-1,3)$

Question 4:

$x^2+ax+b=0 \ \alpha,\beta \ \mbox{are the roots} \\ \\ \Rightarrow \alpha+\beta=-a,\alpha\beta=b \\ \\ \Rightarrow \frac{\alpha+\beta}{\alpha\beta} = \frac{-a}{b} \Rightarrow \frac{1}{\alpha}+\frac{1}{\beta} = \frac{-a}{b} \\ \\ (\alpha+\beta)\left(\frac{1}{\alpha}+\frac{1}{\beta}\right) = \frac{a^2}{b} \\ \\ \Rightarrow \frac{\alpha}{\beta}+\frac{\beta}{\alpha} = \frac{a^2}{b}-2 \\ \\ \mbox{Let the new quadratic be} \ x^2+\frac{B}{A}x+\frac{C}{A}=0 \\ \\ \frac{-B}{A} = \frac{a^2-2b}{b} \ \Rightarrow \frac{B}{A}=\frac{2b-a^2}{b} \\ \\ \mbox{Product of roots ..is obviously 1} \\ \\ \Rightarrow x^2+\frac{2b-a^2}{b} (x) + 1 = 0 \\ \\ \Rightarrow bx^2+(2b-a^2)x+b=0$