anonymous
  • anonymous
Solve by completing the square. x2+6x−16=0 x=__ or x=__
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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calculusxy
  • calculusxy
\(x^2 + 6x - 16 = 0\) \(x^2 + 6x - 16 + 16 = 0 + 16\) \(x^2 + 6x + (\frac{6}{2})^2 = 16 + (\frac{6}{2})^2\) \(x^2 + 6x + 3^2 = 16 + 3^2\) \(x^2 + 6x + 9 = 16 + 9\) \(\sqrt{x^2 + 6x + 9} = \sqrt{25}\) \((x + 3)^2 = -5 \text{ } \text{or} \text{ } 5\) solve for x in these two cases \(x+ 3 = 5\) and \(x + 3 = -5\)
welshfella
  • welshfella
X^2+6x-16=0 (X+3)^2-9-16=0 (X+3)^2 = 25 Can u continue.?
anonymous
  • anonymous
mates got it

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calculusxy
  • calculusxy
what are the two values (just to confirm) ?
anonymous
  • anonymous
you nailed it
anonymous
  • anonymous
good show
calculusxy
  • calculusxy
what is this supposed to be about, may i know ?
anonymous
  • anonymous
solving by completing the squar
anonymous
  • anonymous
thats not right one moment
anonymous
  • anonymous
cumpass says it is squar in a quadratic
anonymous
  • anonymous
square
calculusxy
  • calculusxy
you mean to use the quadratic formula?
anonymous
  • anonymous
yes thank you
calculusxy
  • calculusxy
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
calculusxy
  • calculusxy
and after that just plug those values into the formula: \[x = \frac{ -b \pm \sqrt{b^2 - 4ac} }{ 2a }\]
anonymous
  • anonymous
oh

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