Question

Solve by completing the square.
x2+6x−16=0
x=__ or x=__

Answer 1

\(x^2 + 6x - 16 = 0\)
\(x^2 + 6x - 16 + 16 = 0 + 16\)
\(x^2 + 6x + (\frac{6}{2})^2 = 16 + (\frac{6}{2})^2\)
\(x^2 + 6x + 3^2 = 16 + 3^2\)
\(x^2 + 6x + 9 = 16 + 9\)
\(\sqrt{x^2 + 6x + 9} = \sqrt{25}\)
\((x + 3)^2 = -5 \text{ } \text{or} \text{ } 5\)

solve for x in these two cases
\(x+ 3 = 5\) and \(x + 3 = -5\)

Answer 2

X^2+6x-16=0
(X+3)^2-9-16=0
(X+3)^2 = 25

Can u continue.?

Answer 3

mates got it

Answer 4

what are the two values (just to confirm) ?

Answer 5

you nailed it

Answer 6

good show

Answer 7

what is this supposed to be about, may i know ?

Answer 8

solving by completing the squar

Answer 9

thats not right one moment

Answer 10

cumpass says it is squar in a quadratic

Answer 11

square

Answer 12

you mean to use the quadratic formula?

Answer 13

yes thank you

Answer 14

it's the same concept except you need to know where the a, b, and c values are in a quadratic expression. do you know where they are? hint: ax^2 + bx + c

Answer 15

and after that just plug those values into the formula:
\[x = \frac{ -b \pm \sqrt{b^2 - 4ac} }{ 2a }\]

Answer 16

oh