Answer:
0<x<1
Step-by-step explanation:
(x+3)(x−4)<−12
FOIL
x^2 -4x+3x -12 <-12
x^2 -x -12 <-12
Add 12 to each side
x^2 -x -12+12 <-12+12
x^2 -x <0
Factor out an x
x(x-1) <0
Using the zero product property
x(x-1) =0
x=0 x-1=0
x=0 x=1
Check the ranges
x<0
x(x-1) <0
- * - >0
False
0<x<1
x(x-1) <0
+ * - <0
True
x>1
x(x-1) <0
+ * + >0
False
Hello :
-5^4+4n^2-n+5 : the coefficient of n^2 is : 4<span>
7n^4-5n^3-9n^2+3n+2 : </span><span>he coefficient of n^2 is :-9</span>
Answer:
Step-by-step explanation:
Let the solution to
2x^2 + x -1 =0
x^2+ (1/2)x -(1/2)
are a and b
Hence a + b = -(1/2) ( minus the coefficient of x )
ab = -1/2 (the constant)
A. We want to have an equation where the roots are a +5 and b+5.
Therefore the sum of the roots is (a+5) + (b+5) = a+ b +10 =(-1/2) + 10 =19/2.
The product is (a+5)(b+5) =ab + 5(a+b) + 25 = (-1/2) + 5(-1/2) + 25 = 22.
So the equation is
x^2-(19/2)x + 22 =0
2x^2-19x + 44 =0
B. We want the roots to be 3a and 3b.
Hence (3a) + (3b) = 3(a+b) = 3(-1/2) =-3/2 and
(3a)(3b) = 9(ab) =9(-1/2)=-9/2.
So the equation is
x^2 +(3/2) x -9/2 = 0
2x^2 + 3x -9 =0.
I believe that is the commutative property...which basically states that u can move the numbers around and it will not change the result
