Question

Question #8ce4f

Answer 1

`x=(5+-sqrt(5))/2`

Explanation:

Note that
`color(white)("XXX")color(orange)((x-1)^2)=color(orange)(x^2-2x+1)`

So
`color(orange)((x-1)^2)color(magenta)(-3x+4)=0`
is equivalent to
`color(orange)(x^2-2x+1)color(magenta)(-3x+4)=0`
or
`x^2-5x+5=0`
or
`color(blue)1x^2color(red)(-5)x+color(green)5=0`

Using the quadratic formula:
`color(white)("XXX")`a quadratic of the form `color(blue)ax^2+color(red)bx+color(green)c=0`
`color(white)("XXX")`has solutions `x=(-color(red)b+-sqrt(color(red)(b)- 4color(blue)acolor(green)c))/(2color(blue)a)`

So `color(blue)1x^2color(red)(-5)x+color(green)5=0`

has solutions `x= (+color(red)5+-sqrt((color(red)(-5))^2-4 * color(blue)1 * color(green)5))/(2 * color(blue)1)`

`color(white)("XXXXXXXX")=(5+-sqrt(5))/2`

Answer 2

`x=5/2 +- sqrt5 /2`

Explanation:

Let x-1=y so that the given eq becomes `y^2 -3y-3+4=0`

Or, `y^2 -3y+1=0`

Using the quadratic formula, it would be y=`(3+-sqrt(9-4))/2`

Or, `y=3/2 +- sqrt5 /2`

So that `x=1+3/2 +-sqrt5/2= 5/2 +- sqrt5 /2`