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)

## Comments

equation, not quadraticformula. The second thing, you shall think generally, not in specifics. If I write yx^{2}+zx+s=0, this is still quadratic equation. And m*sin^{2}(x)+h*sin(x)+w=0 is quadratic equation, with respect to sin(x), of course.notan even times an odd !