How do you find the roots, real and imaginary, of $y= (x-3)^2+(x-4)^2$ using the quadratic formula?

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How do you find the roots, real and imaginary, of $y= (x-3)^2+(x-4)^2$ using the quadratic formula?

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Answer 1 · 2016-04-24T20:31:03

$x = 7/2+-1/2i$

Explanation:

First multiply out the quadratic to get it into standard form:

$y = (x-3)^2+(x-4)^2$

$=x^2-6x+9+x^2-8x+16$

$=2x^2-14x+25$

This is now in the form $ax^2+bx+c$ with $a=2$ , $b=-14$ and $c=25$

This has zeros given by the quadratic formula:

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

$=(14+-sqrt((-14)^2-4(2)(25)))/(2*2)$

$=(14+-sqrt(196-200))/4$

$=(14+-sqrt(-4))/4$

$=(14+-sqrt(4)i)/4$

$=7/2+-1/2i$

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How do you find the roots, real and imaginary, of $y= (x-3)^2+(x-4)^2$ 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-3)^2+(x-4)^2 y= (x-3)^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 $y= (x-3)^2+(x-4)^2$ using the quadratic formula?