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

step-by-step explanation:

always plug in x. in this case, x equals 4.

i'm going to first solve 4(2+3x)

1) 4(2+3(4)) use pemdas to solve.

i'm going to multiply whats in the parenthesis first, so 3 times 4=12.

2)4 (2+12) i'm then going to multiply 4 by both numbers in the parenthesis (because multiplying comes before adding)

4 times 2= 8.   4 times 12= 48

4) (8+48)= 56 (answer)

i'm then going to solve 8+12x and plug in 4 for x.

1) 8 +12(4)

multiplying comes before adding, so i'm gonna multiply 12 and 4.

12 times 4= 48.

2) 8+48=56

both equal the same!

is there a graph

step-by-step explanation: