How do you find the roots of #x^2-x=6#?

2 Answers
May 23, 2016

Answer:

#=>x^2-x-6" "=" "(x-3)(x+2)#

Explanation:

Write as #x^2-x-6=0#

Notice that #3xx2 = 6#

And that #3-2=1#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
We need the product (multiplication answer) to be negative (-6)

So either 3 is negative and 2 positive or the other way round as #(-a)xx(+b) = -ab#

But the #-x# as the coefficient of -1

So if #(-a)+(+b) = -1# then #-a# must have the biggest value

So we have to have #(-3)+(+2) = -1" and "(-3)xx(+2)=-6# all as required.

#=>x^2-x-6" "=" "(x-3)(x+2)#

May 23, 2016

Answer:

The solutions/roots to #6=x^2-x# are #x=-2,+3#.

Explanation:

We have
#x^2-x=6#
We need to put this in standard form ( #ax^2+bx+c=y#), we get
#x^2-x-6=0# .
with #a=1#, #b=-1#, and #c=-6# .

You have three ways of solving a quadratic equation:

1) Use the quadratic formula,
#x_{root1}, x_{root2} = -b/{2a} pm {sqrt(b^2 - 4ac)}/{2a}# ,

where #x_{root1}# comes from using the #pm# as subtraction and #x_{root2}# comes from using the #pm# as addition.

2) Factor, for simple equations with #a=1#, for equations with simple integer roots we can find the factors by looking for a two numbers with add to #b# and multiply to #c# (there is a modification to these method used for equations where #ane0#). This numbers are the factors and are used to convert the equation into factored form (or perhaps it's already in factored form). The roots can be found easily from the factored form, by setting each of the two factors to zero and solving for #x_{root}#.

3) Directly solve the equation by first completing the square to get the expression into vertex form, (or perhaps it's already in vertex form?) then solving the resulting equation (any solvable quadratic equation can be directly solved from vertex form, this is how the quadratic formula is proven).

Since these numbers are simple and method 1 is just plug-in and method 3 is rather obscure unless you're already in vertex form (or something close to it), I will use method 2.

We have
#x^2-x-6=0#

we are looking for factors of #-6# which add to #-1#.
We consider

1st try, #6*(-1)=-6#, #-1+6=5# Nope

2nd try, #(-6)*1=-6#, #1-6=-5# Nope

3rd try, #(-2)*3=-6#, #-2+3=1# Nope

4th try, #2*(-3)=-6#, #2-3=-1# Yes!

this means are factors are #(x+2)# and #(x-3)#
our expression becomes
#0=(x+2)*(x-3)#,
(if you expand this expression you will reproduce #0=x^2-x-6#)

We find #x_{root1}# by setting #(x+2)=0#
#x+2=0#
#x=-2#
so #x_{root1}=-2#

We find #x_{root2}# by setting #(x-3)=0#
#x-3=0#
#x=+3#
so #x_{root2}=+3#

The solutions/roots to #6=x^2-x# are #x=-2,+3#.