Rewrite x^2 + 6x - 1 = 0 in the form (x + a) ^2 = k, where a and k are constants. What is k?

answer

10

step-by-step explanation

first find the (x+a) ^2 part

using x^2 + 6x + c, we need to find the value c that completes the perfect square

to do this, divide 6 by two to find a in (x+a) ^2

6/2 = 3 = a

c = a^2

c = 3^2 = 9

plug in values

x^2 + 6x + 9 = (x+3) ^2

compare this to x^2 + 6x - 1, and you can see there is a (9 - - 1) = 10 difference, so subtract 10 from both sides

x^2 + 6x + 9 = (x+3) ^2

x^2 + 6x + 9 - 10 = (x+3) ^2 - 10

x^2 + 6x + - 1 = 0 = (x+3) ^2 - 10

0 = (x+3) ^2 - 10

(x+3) ^2 = 10

k = 10