How do you solve (c - 3) ( c - 1) = 6?

2 Answers
Aug 1, 2017

c=2+-sqrt(7)

Explanation:

First we can put the equation into standard form ax^2+bx+c=0

(c-3)(c-1)=6

<=> Use FOIL to expand the polynomial

c^2-4c+3=6

<=> Subtract 6 from both sides

c^2-4c-3=0

Since you can't factor this polynomial, we can use the quadratic formula

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

In this case

a=1

b=-4

c=-3

and we replace x with the variable c

then we plug in

=> c=(4+-sqrt(16-4(1)(-3)))/2=(4+-sqrt(16+12))/2=(4+-sqrt(28))/2

Since sqrt(ab)=sqrt(a)sqrt(b)

=(4+-sqrt(4)sqrt(7))/2=(4+-2sqrt(7))/2=2+-sqrt(7)

Which means that

c=2+sqrt(7) or c=2-sqrt(7)

Aug 1, 2017

Complete the square to find:

c = 2+sqrt(7)" " or " "c = 2 - sqrt(7)

Explanation:

The difference of squares identity can be written:

a^2-b^2 = (a-b)(a+b)

We can use this with a=c-2 and b=sqrt(7) as follows.

Given:

(c-3)(c-1) = 6

Multiply out the left hand side to get:

c^2-4c+3 = 6

Subtract 6 from both sides to get:

c^2-4c-3 = 0

Complete the square and use the difference of squares identity to find:

0 = c^2-4c-3

color(white)(0) = c^2-4c+4-7

color(white)(0) = (c-2)^2-(sqrt(7))^2

color(white)(0) = ((c-2)-sqrt(7))((c-2)+sqrt(7))

color(white)(0) = (c-2-sqrt(7))(c-2+sqrt(7))

Hence:

c = 2+sqrt(7)" " or " "c = 2 - sqrt(7)