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

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

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Answer 1 · 2016-03-03T23:20:55

Multiply out and simplify to find a quadratic in standard form, then apply the formula to find zeros $x = 3 +-i$

Explanation:

$y=(x-2)^2+(x-4)^2$

$=(x^2-4x+4)+(x^2-8x+16)$

$=2x^2-12x+20$

$=2(x^2-6x+10)$

$x^2-6x+10$ is in the form $ax^2+bx+c$ with $a=1$ , $b=-6$ and $c = 10$ .

This has zeros given by the quadratic formula:

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

$=(6+-sqrt((-6)^2-(4*1*10)))/(2*1)$

$=(6+-sqrt(36-40))/2$

$=(6+-sqrt(-4))/2$

$=(6+-sqrt(4)i)/2$

$=(6+-2i)/2$

$=3+-i$

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

1 Answer

Explanation:

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Science

Math

Humanities

... and beyond

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

1 Answer

Explanation:

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