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

1 Answer
Mar 16, 2016

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.