How do you find the roots, real and imaginary, of #y=(x – 7 )^2-8x+4# using the quadratic formula?

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Answer 1 · 2016-01-25T18:58:20

#x = (22 + 4 sqrt(17))/2 ~~ 19.25#

or

#x = (22 - 4 sqrt(17))/2 ~~ 2.75#

If you would like to use the quadratic formula, you should expand your term first.

#y = (x-7)^2 - 8x + 4#

use the formula #(a-b)^2 = color(green)(a^2) color(brown)(- 2ab) + color(blue)(b^2)#:

#y = color(green)(x^2) color(brown)(- 2 * x * 7) + color(blue)(7^2) - 8x + 4#

#y = x^2 - 14x + 49 - 8x + 4#

#y = x^2 - 22x + 53#

Now, the quadratic formula is

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

and in your case, #a = 1#, #b = -22# and #c = 53#.

Thus, you can apply the formula:

#x = (22 +- sqrt((-22)^2 - 4 * 1 * 53) ) / 2 = (22 +- sqrt(484-212))/2#

# = (22 +- sqrt(272))/2 = (22 +- sqrt(16*17))/2 = (22 +- 4sqrt(17))/2#

You have two solutions:

#x = (22 + 4 sqrt(17))/2 ~~ 19.25#

or

#x = (22 - 4 sqrt(17))/2 ~~ 2.75#