Question

what is the root of x^2-5x-1=0

Answer 1

5.19 , -0,19

Answer 2

\[x^2-5x-1=0\]\[x(x-5)=1\]so\[x=\frac{ 1 }{ 2 }(5-\sqrt{29}) \] and \[x=\frac{ 1 }{ 2 }(5+\sqrt{29}) \]

Answer 3

\[x=\frac{ -5\pm \sqrt{(-5)^{2}-4(-1)} }{ 2 }\]

Answer 4

mhmdrz91 how did you come to that conclusion..i had a hard time following your work..thank you so much too :)

Answer 5

@mhmdrz91 i had a hrad time followng your work

Answer 6

Like @avrillavigne said use the Quad formula and solve it

Answer 7

@avrillavigne im having a hard time following the work, can you explain in detail?

Answer 8

\[ax ^{^{2}}+bx +c=0\]
\[x=\frac{ -b \pm \sqrt{b ^{2}-4ac} }{ 2a }\]

Answer 9

now just compare the formula with the equation u have !

Answer 10

@avrillavigne when i plugged everything into the formula i wound up with
\[x=\frac{ -5x \sqrt{5x2-4x^2-1} }{ 2x^2 }\]

Answer 11

there shouldn't be any x's on the right side!
look closely to the formula

Answer 12

i dont see it @avrillavigne, i dont see how i lose the X's when i plug them in if a=x^2 and B = -5x then thats what i wind up with

Answer 13

u see a,b,c are constants!
for example in your equation a=1 , b=-5 ,c=-1

Answer 14

so i can just remove the X's? ooo ok let me see what i get

Answer 15

there u go ! I think now u get it

Answer 16

well i got it somewhat i got \[x=\frac{ -5\sqrt{29} }{ 2 } when \in the answer \it shhould be a positive 5\]

Answer 17

sorry about the scribble in the end...it should say the answer should be a poistive 5 instead of a negative 5

Answer 18

how about this \[\frac{ -(-5)\pm \sqrt{29} }{ 2 }\]

Answer 19

oooo thank you so much i think i got it now! :)

Answer 20

good luck