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

2 Answers
Lotusbluete
Feb 2, 2016

#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)#

Veddesh #phi#
Feb 4, 2016

#(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#

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