How do you solve $x^2 - 6x + 9 = 0$ using the quadratic formula?

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Answer 1 · 2016-07-10T21:38:50

x = 3

$x^2 - 6x + 9 = (x - 3)^2$
There ia a double-root at x = 3

Tony B

Using the formula as instructed.

Compare to: $ax^2+bx+c -> x=(-b+-sqrt(b^2-4ac))/(2a)$

In this case: $a=1; b=-6 and c=9$ giving:

$x=(+6+-sqrt((-6)^2-4(1)(9)))/(2(1))$

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

$x=3+-0$

$x=3$

How do you solve x^2 - 6x + 9 = 0x^2 - 6x + 9 = 0 using the quadratic formula?