Question

Solve by using quadratic formula
r^2+2r=1

Answer 1

subtract 1 from both sides to get r^2+2r-1=0

Compare your quadratic equation with \(ar^2+br+c=0\)
find \[a=...?\\b=...?\\c=...?\]
then the two roots of r are:
\(\huge{r_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}\)

Answer 2

a=1 b=2 c=-1 right?

Answer 3

correct :)
now just plug it in the formula

Answer 4

Hmm i can't type the equation where i'm stuck

Answer 5

I'm only using mobile, it's hard to type the equation

Answer 6

lets go step by step,
\(b^2-4ac=...?\)
just give final values..

Answer 7

8

Answer 8

good, so numerator = \(-2 \pm \sqrt 8\)
right ?
can you simplify \(\sqrt 8\) ??

Answer 9

2 square root of 2

Answer 10

I can't use symbols sorry

Answer 11

i understood that , its \(2 \sqrt2\)and is correct :)
so, \(\huge{x_{1,2}=\frac{-2 \pm 2\sqrt2}{2} = -1 \pm \sqrt2}\)
so the 2 roots are
-1+ root 2, and -1- root 2

Answer 12

I can't see the whole equation

Answer 13

r1,2 = (-2 +or - 2sqrt2 )/2=-1 + or - sqrt 2
so, 2 roots are
r1 = -1+sqrt 2
r2 =-1-sqrt2

Answer 14

How did it become -1?

Answer 15

factoring out 2 in the numerator
(-2 +or - 2sqrt2 )/2
= 2 (-1 +or - sqrt 2)/2
the 2 in the numerator and denominator cancels out

Answer 16

Oh i see, sorey, i get it :£

Answer 17

:)

Answer 18

great ! :)