A. that is b^2-4ac
for
ax^2+bx+c
the deterimant (part under the radicand)
is b^2-4ac
if it is
1. less than 0 then there are no real solutions
2. eqal to 0 then there is 1 real solution
3. greater than 0, there are 2 real roots
9x^2-16x+60
b^2-4ac=(-16)^2-4*9*60=256-2160<0
no real solutions
B. grouping
ac method
4 times -5=-20
what 2 numbers multiply to get -20 and add to get 8?
-2 and 10
seperate the middle number to get that
4x^2-2x+10x-5=0
(4x^2-2x)+(10x-5)=0
2x(2x-1)+2(2x-1)=0
(2x-1)(2x+2)=0
set equal to 0
2x-1=0
2x=1
x=1/2
2x+2=0
2x=-2
x=-1
x=1/2 or -1
3x-5=-15 2x-7=-14 there you go
Answer: -1 three times and -1 five times
Step-by-step explanation:
-1 three times = -1 X -1 X -1 = -1
-1 two times = -1 X -1 = 1
-1 four times = -1 x -1 x -1 x -1 = 1
-1 five time = -1 x -1 x -1 x -1 x -1 = -1
Therefore -1 three times and -1 five times will be negative
(2.5)^2 + (16.5)^2 = c^2
c = sqrt(2.5^2 + 16.5^2)
c = sqrt(278.5)
c = 16.69
Rounded up it would be 17, so the answer is D.
Answer:
-2x/7-x
Step-by-step explanation:
(4x-42)/7-x + 6/1 (7-x)=(4x-42+42-6x)/7-x=-2x/7-x
