anonymous
  • anonymous
Solve x2 − 3x = −8
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
@Michele_Laino
pooja195
  • pooja195
a ,b , c values are \[\huge\rm Ax^2+Bx+C=0\] where a =leading coefficient b= leading coefficient c= constant term So we need to add 8 to both sides \[\huge~\rm~ x^2 − 3x+8 \] Now factor from here What numbers multiply to make 8 but add to make -3? if there are none the answer would be prime
anonymous
  • anonymous
@ganeshie8 @Hero

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More answers

pooja195
  • pooja195
?
anonymous
  • anonymous
idk what numbers multiply to 8 and add to -3
anonymous
  • anonymous
@pooja195
pooja195
  • pooja195
Then clearly there is not solution :) reread what i wrote are you given options?
anonymous
  • anonymous
no solution isnt an option
pooja195
  • pooja195
What are the options?
anonymous
  • anonymous
Um I don't know how to right them out
pooja195
  • pooja195
Just type them out
pooja195
  • pooja195
Or scrnshot
anonymous
  • anonymous
okay hold on
anonymous
  • anonymous
1 Attachment
pooja195
  • pooja195
I havent learned "i" yet.... @UsukiDoll
anonymous
  • anonymous
Use the quadratic formula.
UsukiDoll
  • UsukiDoll
HUH?! O_O! that was random
UsukiDoll
  • UsukiDoll
yeah use quadratics on this equation... use discriminant formula b^2-4ac to see if we have a perfect square or not perfect square - we can factor not a perfect square - use quadratic formula
anonymous
  • anonymous
wait what do i do?
UsukiDoll
  • UsukiDoll
\[\huge~\rm~ x^2 − 3x+8 \] where a = 1, b = -3, and c = 8 plug it into the discriminant formula which is b^2-4ac \[\LARGE (-3)^2-4(1)(8) \]
anonymous
  • anonymous
okay what next?
UsukiDoll
  • UsukiDoll
\[\LARGE (-3)^2-4(1)(8) \] \[\LARGE 9-32 = -23 \] not a perfect square and a negative... ok looks like we're going to have complex roots.
UsukiDoll
  • UsukiDoll
now we solved the b^2-4ac already so that makes the quadratic formula a bit easier \[\LARGE \frac{ -b \pm \sqrt{b^2-4ac}}{2a}\]
UsukiDoll
  • UsukiDoll
since a = 1, b = -3, and c = 8 \[\LARGE \frac{ -(-3) \pm \sqrt{-23}}{2}\] negatives aren't allowed in the radical...so we write an i which stands for imaginary. \[\LARGE \frac{ 3 \pm \sqrt{23}i}{2}\]
anonymous
  • anonymous
so the answer would be b?
UsukiDoll
  • UsukiDoll
wait... I haven't looked at that portion yet.
UsukiDoll
  • UsukiDoll
.docx hates me :(
anonymous
  • anonymous
okay ill try to get it on something else
UsukiDoll
  • UsukiDoll
finally got through to it.. it is the second choice.
anonymous
  • anonymous
Thank you!!!!

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