x2+3x-10=0?

Question

A.-2,5
B-5,-2
C.2,5
D.-5,2

Answer 1

Answer D.
x²+3x-10=0
(x+5)(x-2) = 0
x = 5 or x = -2
d = 3² - 4*1*-10
d = 9 + 40
d = 49 ==> d > 0
There are two different real roots.
x = (-b +/- \/d) : 2a
x = (-3 +/- \/49) : 2
x' = (-3 +/- 7) : 2
x' = (-3 + 7) : 2 = 4: 2 = 2
x" = (-3 - 7) : 2 = -10 : 2 = -5
Solution: {x belongs to R| x' = 2 and x" = -5}
><><

Answer 2

OK. First, you have to factor this expression. You have x^2, so the equation looks like this:

(x+?)(x+?) Now you have to find a number that multiplies to make -10, and adds up to make 3. This is factoring. Those numbers are 5 and -2. So now, you equation looks like this:

(x+5)(x-2) Now to find the =0 part of the problem, you have to find the number in each parenthesis that will equal 0. Those numbers are -5 because -5 plus 5 equals 0 and 2 because -2 +2 equals 0.

This means the answer is D.

This will work for any other problem similar to this. I hope I helped! Good Luck!!!

Answer 3

D
Factor X^2+3x-10=(X-2)(X+5)
(X+5)=0 if X=-5
(X-2)=0 if X=2

Answer 4

X^2-3x-10=0 (X-5) (X+2)=0 -2, 5

Answer 5

A: (x + -2)(x +5)

Answer 6

this is the Zero Product Property, you must break it down, or somthing
You must make x=0
x2+3x-10=0
10
3

Answer 7

assuming that's a square,
it would factor into (x + 5)(x - 2)
so... what would x be in order for either of those to equal 0?
D, of course

Answer 8

(x+5)(x-2) = 0
x = -5,2

Answer 9

(x+5)(x-2)

x = -5 ro 2

D

Answer 10

D