Question

r^2-22r+21+0

Answer 1

you mean

r^2-22r+21=0

Answer 2

yes

Answer 3

or
r^2-22r+121=0
?

Answer 4

the first one you suggested

Answer 5

ok so find the factors of 21

\[\pm1,\pm3,\pm7,\pm21\]

which of these add to -22 ?

Answer 6

-1,-21?

Answer 7

right,

Answer 8

r^2-22r+21+0
r^2-22r+21

a = 1, b = -22, c = 21;
r = (-b +- sqrt(b^2 -4ac))/2a
r = (-(-22) +- sqrt(484- (-84)))/2
r = (22 +- sqrt(568))/2


for (+)

r = (22+ 23.8)/2 = 22.9

for (-)

r=(22 - 23.8)/2 = -0.9

Answer 9

(r-1)(r-21) ?

Answer 10

yes

Answer 11

(r-1)(r-21)

where r =1 or 21

Answer 12

to check , expand the brackets

(r-1)(r-21)= r(r-21)-1(r-21)
= r^2-21r-r+21
= r^2-22r+21

Answer 13

good

Answer 14

thank you very much! both of you! i think i actually get it now

Answer 15

ok ur welcome., friend, follow my twitter ok @gerryphysicist

Answer 16

gerryliyana

you mean
r = (-(-22) +- sqrt(484- 84))/2

Answer 17

yes @UnkleRhaukus .., hehe
I am wrong count before

Answer 18

\[r = \frac{-(-22) \pm \sqrt{484- 84}}2\]
\[\quad= \frac{22 \pm \sqrt{400}}2\]\[\quad= \frac{22 \pm 20}2\]\[\quad=11 \pm 10\]\[r_{1,2}=21,1\]