Question

How do you solve `(y-3)^2=4y-12`?

Answer 1

`y=7`

Explanation:

`(y-3)^2=4y-12`
or
`(y-3)^2=4(y-3)`
or
`y-3=4`
or
`y=4+3`
or
`y=7`

Answer 2

`y=+3" and "y=+7` are solutions

Explanation:

This is a quadratic in `y`

That is: instead of `y=ax^2+bx+c` we have `x=ay^2+by+c`

This has the effect of rotation the standard graph of `uu` to `sub`
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Given:`" "(y-3)^2=4y-12`

Write as:`" "(y-3)^2-4y+12=0`

Set as equal to `x` giving:

`x=(y-3)^2-4y+12`

Squaring the bracket

`x=y^2-6y+9-4y+12 color(red)(" "larr" "-12" corrected to "+12)`

`x=y^2-10y-21`

Note that `3xx7=21 " and that "3+7=10`

Factorising

`x=0=(y-3)(y-7)`

`y=+3" and "y=+7` are solutions

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