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

Question

1748 views around the world

You can reuse this answer
Creative Commons License

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

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

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