Question

How do you solve `(x-3)^2-18=0`?

Answer 1

`x = 3 +- 3 sqrt(2)`

Explanation:

The answer that was given above is absolutely correct. However, since the equation is given in the form

`(x - a)^2 + b = 0`,

which is basically an intermediate step of the "completion of the circle" method, I would like to show you an easier way to solve it.

`color(white)(x)`

`(x - 3)^2 - 18 = 0`

... add `18` on both sides...

`<=> (x-3)^2 = 18`

Now, you can draw the root, but beware: there are two solutions, the negative and the positive one. (Please be sure that this is clear for you. As an example, for `x^2 = 49`, both `x = 7` and `x = -7` are solutions.)

`=> x - 3 = sqrt(18) " or " x - 3 = -sqrt(18)`

... add `3` on both sides...

`=> x = 3 + sqrt(18) " or " x = 3 - sqrt(18)`

So, your solution is

`x = 3 +- sqrt(18) = 3 +- sqrt(9 * 2) = 3 +- sqrt(9) * sqrt(2) = 3 +- 3 sqrt(2)`

Answer 2

`(x-3)^2-18=0`

Use the formula `(a+b)^2=a^2+2ab+b^2`

`rarr(x-3)^2=x^2+2(x)(-3)+(-3)^2=x^2-6x+9`

So,

`rarrx^2-6x+9-18=0`

`rarrx^2-6x-9=0`

Now this is a Quadratic equation (in form `ax^2+bx^2+c`)

Use quadratic formula:

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

In this case `a=1,b=-6,c=-9`

`rarrx=(-(-6)+-sqrt((-6)^2-4(1)(-9)))/(2(1))`

`rarrx=(6+-sqrt(36-(-36)))/2`

`rarrx=(6+-sqrt(36+36))/2`

`rarrx=(6+-sqrt72)/2`

`rarrx=(6+-sqrt(36*2))/2`

`rarrx=(6+-6sqrt2)/2=6/2+-(6sqrt2)/2`

`rarrx=3+-3sqrt2`