Question

Write x2 − 6x + 7 = 0 in the form (x − a)2 = b, where a and b are integers.

(x − 4)2 = 3
(x − 3)2 = 2
(x − 2)2 = 1
(x − 1)2 = 4

Answer 1

You know the completing the square method?

Answer 2

thats what im confused on

Answer 3

as is, it will not factor to that form, it can be (x-6)*(x-1)=0, but they want that
(x-a)^2 = b

...

Answer 4

x^2 - 6x + 7 =0

move 7 over...
\[x^2 - 6x = -7\]

Answer 5

To complete the square on that left side, you can add to both sides of the equation, half of that b term squared... in
ax^2 + bx + c
\[\large (\frac{ b }{ 2 })^2\]

Answer 6

i can show you where that comes from if you want, or you can just remember that..

Answer 7

isnt that using completing the square and using the vertex or something?

Answer 8

nvm i really dont know what im talking about XD

Answer 9

yeah if you take that half of b squared and add it on to the thing, it will become a square that factors to (x - a)^2

\[\large a*x^2+b*x\]
\[\large ax^2 + bx + (\frac{ b }{ 2 })^2\]

Answer 10

you have a -6 there, so half of that squared is \[\large (\frac{ -6 }{ 2 })^2 = 9\]

x^2 - 6x = -7
add that to both sides of the equation

\[\large x^2 - 6x + 9 = -7 + 9\]

Answer 11

a=1
b=-6
c=7
right?
now i just plug it in?

Answer 12

oh you already did it

Answer 13

As is it is not a perfect square, first move the 7 over to the other side to get

x^2 - 6x = -7

now add on half of (-6) squared, or 9 to both sides

x^2 - 6x + 9 = -7 + 9
\[\large x^2 - 6x + 9 = 2\]
now it is a square that will factor

\[\large (x - 3)^2 = 2\]

Answer 14

ohhhh ok

Answer 15

thanks for helping^

Answer 16

welcome, yeah just remember half b squared , to add on and complete the square..

Answer 17

alright:)