How do you solve (c−3)(c−1)=6?

2 Answers
256
Aug 1, 2017

c=2±√7

Explanation:

First we can put the equation into standard form ax2+bx+c=0

(c−3)(c−1)=6

⇔ Use FOIL to expand the polynomial

c2−4c+3=6

⇔ Subtract 6 from both sides

c2−4c−3=0

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

x=−b±√b2−4ac2a

In this case

a=1

b=−4

c=−3

and we replace x with the variable c

then we plug in

⇒c=4±√16−4(1)(−3)2=4±√16+122=4±√282

Since √ab=√a√b

=4±√4√72=4±2√72=2±√7

Which means that

c=2+√7orc=2−√7

George C.
Aug 1, 2017

Complete the square to find:

c=2+√7 or c=2−√7

Explanation:

The difference of squares identity can be written:

a2−b2=(a−b)(a+b)

We can use this with a=c−2 and b=√7 as follows.

Given:

(c−3)(c−1)=6

Multiply out the left hand side to get:

c2−4c+3=6

Subtract 6 from both sides to get:

c2−4c−3=0

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

0=c2−4c−3

0=c2−4c+4−7

0=(c−2)2−(√7)2

0=((c−2)−√7)((c−2)+√7)

0=(c−2−√7)(c−2+√7)

Hence:

c=2+√7 or c=2−√7

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