Question

x^2-5x+1+0

Answer 1

x^2-5x+1=0

Answer 2

Are you meant to find x or just factorise?
If you are to find x then follow this.
\[x^2-5x+1=0\]
Use the quadratic formula.
\[x=\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\]

Answer 3

it says to find all real solutions of the equation by completing the square

Answer 4

ah okay, then don't follow what I did there.

Answer 5

\[x^2-5x=-1\]
\[(x-(\frac{ 5 }{ 2 }))=-1+(\frac{ 5 }{ 2 })^{2}\]

Answer 6

To complete the square, move the constant to the other side.
\[x^2-ax\]
\[(x-\frac{a}{2})^{2}\]

Answer 7

I did a mistake in the second line.

Answer 8

meant to be
\[(x-\frac{ 5 }{ 2 })^2=-1+(\frac{ 5 }{ 2 })^2\]

Answer 9

I had it x^2-5x- 25/4= -1 then moved the 25/4 to the other side

Answer 10

That's not how you complete the square.

Answer 11

so I was at x^2-5x= -1 + 6 1/4 which equaled 5 1/4

Answer 12

You're not meant to expand the LHS.

Answer 13

oh ok

Answer 14

\[(x-\frac{ 5 }{ 2 })^2=\frac{ 21 }{ 4 }\]
\[x-\frac{ 5 }{ 2 }=\frac{ \pm \sqrt{21} }{ 2 }\]

Answer 15

Move the -5/2 to the RHS.

Answer 16

So then it would be
\[x=\frac{ \pm \sqrt{21} }{ 2 }+\frac{ 5 }{ 2 }\]
\[x=\frac{ 5\pm \sqrt{21} }{ 2 }\]

Answer 17

thanks for the help

Answer 18

No worries.