Question

How do you find the roots, real and imaginary, of `y= x^2-6x+(x-4)^2` using the quadratic formula?

Answer 1

real roots: `x=(7+-sqrt(17))/2`
imaginary roorts: none

Explanation:

`1` Start by converting the equation into standard form.

`y=x^2-6x+(x-4)^2`

`y=x^2-6x+(color(crimson)x` `color(blue)(-4))(color(orange)x` `color(teal)(-4))`

`y=x^2-6x+(color(crimson)x(color(orange)x)` `color(crimson)(+x)(color(teal)(-4))` `color(blue)(-4)(color(orange)x)` `color(blue)(-4)(color(teal)(-4)))`

`y=x^2-6x+(x^2-4x-4x+16)`

`y=x^2-6x+(x^2-8x+16)`

`y=x^2+x^2-6x-8x+16`

`y=color(brown)2x^2` `color(turquoise)(-14)x` `color(violet)(+16)`

`2`. Identify the `color(brown)a,color(turquoise)b,` and `color(violet)c` values of the quadratic equation. Then plug the values into the quadratic formula to solve for the roots.

`color(brown)(a=2)color(white)(XXXXX)color(turquoise)(b=-14)color(white)(XXXXX)color(violet)(c=16)`

`x=(-b+-sqrt(b^2-4ac))/(2a)`

`x=(-(color(turquoise)(-14))+-sqrt((color(turquoise)(-14))^2-4(color(brown)2)(color(violet)(16))))/(2(color(brown)2))`

`x=(14+-sqrt(196-128))/4`

`x=(14+-sqrt(68))/4`

`x=(14+-2sqrt(17))/4`

`x=(2(7+-sqrt(17)))/(2(2))`

`x=(color(red)cancelcolor(black)2(7+-sqrt(17)))/(color(red)cancelcolor(black)2(2))`

`color(green)(|bar(ul(color(white)(a/a)x=(7+-sqrt(17))/2color(white)(a/a)|)))`

`:.`, the real roots are `x=(7+-sqrt(17))/2`.