soumya1989 wrote:
If \(x^2 +2x -15 = -m\), where x is an integer from -10 to 10,inclusive, what is the probability that m is greater than 0?
A. 2/7
B. 1/3
C. 7/20
D. 2/5
E. 3/7
Two ways..
(I) Also explained above..
\(x^2+2x-15=-m.....x^2+5x-3x-15=-m........(x+5)(x-3)=-m\)
Now x can take any value from -10 to 10, but m >0.
When x is -5 or 3, m is 0 as roots of quadratic equation \((x+5)(x-3)=0\) are -5 and 3.
Now if we take any value less than -5, both (x+5) and (x-3) will be negative and their product will be positive that is -m>0...m<0
Similarly, if we take any value more than 3, both (x+5) and (x-3) will be positive and their product will be positive that is -m>0...m<0
However for values between -5 and 3, (x+5) will be positive and (x-3) will be negative and their product will be negative.. Thus -m<0....m>0
So, the values -4, -3, -2, -1, 0, 1, 2 will fit in..
so 7 values out of 21 values will give a probability of \(\frac{7}{21}=\frac{1}{3}\)
(II) Next could be
\(x^2+2x-15=-m.....x^2+2x=15-m........(x)(x+2)=15-m......m=15-x(x+2)\)
So, sum of two consecutive even integers should be less than 15 for m>0..
Now, -5*-3 = 15 so x>-5 AND 3*5=15, so x<3
-5<x<3... Rest same as above for probability
B