Question

solve quadratic inequalities: x2-2x is greater than or equals to 0

Answer 1

are you missing a ^ sign , i.e. is it x squared - 2*x

Answer 2

Oh its x^2-2x>=0 right?

Answer 3

Can u convert x^2-2x into completed square form

Answer 4

its x^2-2x>=0

Answer 5

find the zeros. it is negative between them and positive outside them because it is a parabola that opens up

Answer 6

x^2-2x>=0
x^2-2*x*1+1^2>=1^2
(x-1)^2>=1

Answer 7

hmmm

Answer 8

Now, HINT: IF A^2>=(+-b)^2 then A<=-b AND A>=b

Answer 9

lets try this
\[x^2-2x\geq0\]
\[x(x-2)\geq 0\] zeros are at \(x=0\) and at \(x=2\)
it will be negative between the zeros, i.e. on the interval \((0,2)\) and positive outside the zeros, on the intervals
\[(-\infty,0)\cup (2,\infty)\]

Answer 10

i dont understand

Answer 11

\(y=x^2-2x\) is a parabola that open up

Answer 12

it is negative (below the \(x\) axis) between the zeros
positive (above the \(x\) axis) outside the zeros

Answer 13

@satellite73 thank you

Answer 14

This may also help:
x^2-2x>=0
x^2-2*x*1+1^2>=1^2
(x-1)^2>=1
(x-1)^2>=(+-1)^2
Now,
x-1<=-1 AND x-1>=1
x<=0 AND x>=2

Answer 15

not to be picky, but there is no such thing as \(x\leq 0\) AND \(x\geq 2\) it should be
\[ x\leq 0 \text{ or } x\geq 2\]