anonymous
  • anonymous
Solve the quadratic equation y2 - 5y = -4 by factoring. {-2, 2} {2, 2} {-1, 4} {4, 1} i factor it, but i aint getting any of these?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
@stacy.reed08
anonymous
  • anonymous
@Nnesha
Nnesha
  • Nnesha
show ur work so i can find out the mistakes otherwise i have to start it ovr

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anonymous
  • anonymous
|dw:1439742404806:dw| thats all i got to,
Nnesha
  • Nnesha
cool now find two numbers when you multiply them you should get product of AC and when you add or subtract them you should get the middle term
anonymous
  • anonymous
i did, i tried all the option and i couldnt find anything...
anonymous
  • anonymous
i for sure thought it would be like (-1,4) but i didnt get 5
Nnesha
  • Nnesha
no i mean a-leading coefficient c=constant term |dw:1439742684150:dw| \[Ax^2+Bx+C=0\]
Nnesha
  • Nnesha
what two number you should multiply to get positive 4 but when you add them you should get -5
anonymous
  • anonymous
-4 and -1
Nnesha
  • Nnesha
yes right so|dw:1439742854487:dw| set both parentheses equal to zero then solve for x
Nnesha
  • Nnesha
i mean for y*
anonymous
  • anonymous
y=4 y=1
Nnesha
  • Nnesha
yep
anonymous
  • anonymous
i was thinking it need to be (-4,-1) like just the numbers used to factor
Nnesha
  • Nnesha
no they are asking *solutions ) so you have to solve for the variable :)
anonymous
  • anonymous
i would have like it to say *factor completely* bc when i was in public school, thats what it ment
anonymous
  • anonymous
\[y^2-5y=-4\]\[y^2-5y+4=-4+4\]\[y^2-5y+4=0\]\[y^2-y-4y+4=0\]\[y(y-1)-4(y-1)=0\]\[(y-4)(y-1)=0\]and next the very last step...

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