Can you give me solution for this. I'm unable to understand this concept.
TIA

soumya, what you have shown in your solution is not correct, if m is positive, then x^2 + 2x -15 will be negative according to the equation. So for m to be positive, x^2 +2x -15 has to be negative, i.e x^2 +2x -15 < 0. After factorizing the left side, it comes as (x+5)(x-3)<0. After solving this inequality, -5<x<3, so x can be 7 values. And probability of this x out of 21 is 7/21, i.e the answer is 1/3 (B).

The answer is (B).

There could be 21 values of x ranging from -10 to 10.The values of x i.e. -4,-3,-2,-1,0,1,2 (total 7 values of x) could result m greater than zero. Hence , the probability is 7/21 or 1/3.

Think of it this way
first, there are 21 numbers between -10 and 10, since 0 is included, if youre not sure why, use the arithmetic series formula to be sure
10= -10 +1(n-1)
20=n-1
21 = n

that being taken care of...

we have x^2+2x-15= -m


they ask what is the change that m is greater than 0.
if thats the case, then -m is a negative number, which is less than zero,
rewrite the inequality like this.

x^2+2x-15 <0

factor

(x-3)(x+5)<0

the solutions we see are 3 and -5
that means the solution is between 3 and negative 5, non inclusive.

then the solutions can be any integer between 3 and -5, which are 2,1,0,-1,-2,-3,-4.
thats seven numbers out of 21 possible solutions. so 7/21= 1/3

the answer is B

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

Bump for further discussion

can you elaborate

Please, refer to the explanations above .

Regards

Have a look at the graph way, maybe it'll help.

We know the roots are -5 and 3 and that's where the graph becomes 0. Also, we have to find the values where y <0. So, that's 7 integer points.
P = 7/21 (-10 to 10) = 1/3 (B)

x^2 + 2x - 15 = (x+5)(x-3)
If m > 0, then -m < 0.
Question stem, rephrased:
What is the probability that (x+5)(x-3) < 0 ?

We can use the CRITICAL POINT APPROACH.

Critical points occur when the two sides of the inequality are EQUAL.
In the inequality above, the left side is equal to 0 when x=-5 or x=3.
To determine which ranges satisfy the inequality, test one value to the left and one value to the right of each critical point.
Here, we must test x<-5, -5<x<3 and x>3.
If we test x=-10, x=0, and x=10, only x=0 satisfies (x+5)(x-3) < 0.
Implication:
-5<x<3 is the only valid range.

Thus:
Of the 21 integers between -10 and 10, inclusive, only the 7 integers between between -5 and 3
satisfy (x+5)(x-3) < 0, yielding the following probability:
7/21 = 1/3

For m to be positive equation becomes

x^2 + 2x - 15 < 0

(x-3)(x+5) <0

x-3 <0

x<3


x + 5 >0

x > -5


-5 <x<3

required range = -4, -3, -2, -1, 0, 1, 2

Total possibilities 10-(-10) = 20+1 = 21


Therefore 7/21

1/3


Answer B

Adewale Fasipe, GRE and GMAT quant instructor from Nigeria.