Question

16r2-8r=-1

Answer 1

you have the quadratic equation there:

\[\LARGE 16r^2-8r+1=0\]
try its formula:
\[\LARGE {r_{1/2}} = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
and see what you get ...

Answer 2

i want the answer

Answer 3

well I won't give the answer to you, if you don't try !!
where are you stuck ?

Answer 4

why

Answer 5

\[\LARGE {r_{1/2}} = \frac{{ - ( - 8) \pm \sqrt {64 - 4 \cdot 16 \cdot 1} }}{{2 \cdot 16}}\]
tell me what you get...

Answer 6

30

Answer 7

try again... here's a hint.
\[\LARGE {r_{1/2}} = \frac{{8 \pm \sqrt {64 - 64} }}{{32}}\]

Answer 8

x=8+32

Answer 9

64-64=?

Answer 10

0

Answer 11

ok so you have:
\[{r_{1/2}} = \frac{{8 \pm \sqrt 0 }}{{32}}\]
\[{r_{1}} = \frac{{8 }}{{32}}\]
and
\[{r_{2}} = \frac{{8 }}{{32}}\]
what do you get now ? :)

Answer 12

1/4

Answer 13

that's correct. Well done

Answer 14

x2+5x-6

Answer 15

what's that?

Answer 16

(x+6)(x-1)=0

x=-6 x=1