How do you find the roots, real and imaginary, of y= x^2-6x+(x-4)^2 using the quadratic formula?
1 Answer
real roots:
imaginary roorts: none
Explanation:
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)
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)|)))