please answer and please explain...?

Question

For what values of x is x^2+2x-15>0?

Answer 1

Break down the equation in its simplest form, you get

(X + 5) (X-3)

Now, assume that the equation is equivalent to zero.

x = -5
x = 3

Any value greater than 3 (i.e. 4, 5, 6, etc..) and any value lesser than -5 (-6, -7, -8, etc.)

Try to substitute any of the above example values and you would get an answer which is greater than zero in the equation.

Answer 2

x^2 + 2x - 15 > 0

In order to solve this problem, first factor.

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

At this point, you solve it as if it were an equation, in order to get the critical numbers.

(x - 3) (x + 5) = 0 implies x = {3, -5}

Therefore, your critical numbers are 3 and -5.

At this point, we want to TEST the behaviour around them, for positivity and negative. On a number line, we're going to have three regions: x < -5, x is between -5 and 3, and x is greater than 3.

The convenient thing is that we only have to test ONE value within this region, because if one value is negative or positive, it follows that all numbers in that region will be negative/positive in that region too.

Since the expression is greater than 0, we want to find positive numbers.

For x < -5, test x = -6. Then
(-6)^2 + 2(-6) - 15 = 36 - 12 - 15 = 24 - 15 = 9, which is positive.
Include the interval (-infinity, -5)

For -5 < x < 3, test 0. Then
0^2 + 2(0) - 15 = -15, which is negative. We don't care.

For x > 3, test 10. Then
10^2 + 2(10) - 15 = 100 + 20 - 15 = 105, which is positive.
Include the interval (3, infinity)

Therefore, our solution set is

x is an element of (-infinity, -5) U (3, infinity)

The U means union.

Answer 3

Let x^2+2x-15=0
Then,(x+5)(x-3)=0 [factorising x^2=2x-15=0]
Therefore x=-5 or 3
Therfore,for any value of x less than -5 or greater than 3,the value of the expression would be greater than 0.

Answer 4

First, factor the polynomial

x^2 + 2x - 15 = (x+5)(x-3)

Now, set each of those terms =0 and solve for x

x+5=0
x=-5

x-3=0
x=3

Now, to solve, pick numbers on either side of those to get the solution:

if x=-6.... (-6)^2 +2(-6)-15 = 9>0
so, x<-5

if x=-4... (-4)^2+2(-4)-15 = -7<0
so, x can't be between -5 and 3.

if x=4... (4)^2 +2(4)-15 = 9>0
so, x>3

You solution is {x<-5 or x>3}

Answer 5

take this formula and substitute the values and plz simplify on ur own ..........its kinda lengthy...........
x= -b +/- sq.root of b^2-4ac
the whole thing divide by 2a
now i will telll u the values..............
-2 +/- sq.root 2^2- (4)(1)(-15)
/by 2(1)
= -2 +/- sq.root 4 - (-60)
/by2
= -2 +/- sq.root 4 +60
/by 2
= -2 +/- sq root of 64
/by 2
= -2 +/- 8
/by 2
= (-2+8) / 2 = 3
AND
= (-2 - 8 ) / 2 = -5
SO: x+3=0 OR x - 5 = 0 (we take both the digits to the other side)
then = x = -3 Or x = 5
hope it works

Answer 6

x^2+2x-15>0
(x+5)(x-3)>0
there are 2 cases
1case
(x+5)<0 and (x-3)<0
x<-5 and x<3
so x<-5

or

2case
(x+5)>0 and (x-3)>0
x>-5 and x>3
so x>3

so answer is union of both the cases
ie
3

Answer 7

im not going to do ur homework for you..