Solving quadratic functions with equations
Hi Raven,
2n^2 + 4n -16 = 0
This is standard form aN^2 + bN + c (a=2,b=4,c= -16)
1)examine the possible factors of the a & c constants [ c=(-2)(8) or (2)(-8) or (4)(-4) or (1)(-16) or (-1)(16) ]
[a = (1)(2) ]
2)look at the possible combinations and try to construct the 'b' constant from the candidate factors. For example consider the form
(2n + Q)(n + R)
QR = c = -16
(Q + 2R) = b = 4
After a little study, you'll see that (Q=8 , R=-2) yields the correct 'b' So we FOUND the solution just by examination. This doesn't work all the time, but is faster if the roots are simple
(2n + 8)(n - 2)-----> n=-4, n=2 are the solutions
2)If you cannot find simple factors, you are left with the quadratic formula which always provides the correct answer: n = [ -b ± sqrt(b^2 - 4ac) ]/2a
n= [ -4 ± sqrt(16 - 4(2)(-16) ]/2(2)
n= -1 ± sqrt(16 + 128)/4
n= -1 ± sqrt(144)/4 (sqrt(144)=12)
n= -1 ± 3
n= -4, 2 (Same as we got above)
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