Solving quadratic functions with equations

## 2n^2+4n-16=0

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# 3 Answers

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)

Hi Raven;

2n

^{2}+4n-16=0For the FOIL...

FIRST must be (2x)(n)=2n

^{2}.OUTER and INNER must add-up to 4n.

LAST must be (16)(1) or (1)(16) or (8)(2) or (2)(8) or (4)(4) and one number must be negative to produce -16.

(2n+8)(n-2)=0

Let's FOIL...

FIRST...(2n)(n)=2n

^{2}OUTER...(2n)(-2)=-4n

INNER...(8)(n)=8n

LAST...(8)(-2)=-16

2n

^{2}-4n+8n-16=02n

^{2}+4n-16=0(2n+8)(n-2)=0

Either or both parenthetical equations must equal zero.

2n+8=0

2n=-8

n=-4

n-2=0

n=2

First, divide both sides by 2. You will get:

n

^{2}+2n-8=0Now either factor out or use quadratic formula.

1) Factoring.

Product of roots shall be equal to -8. Sum of them equals to -2. Obvious choice is -4 and 2. (-4)*2=-8, -4+2=-2.

So we have the following factoring: (n+4)(n-2)=0. This means that either n=2 or n=-4.

2) Quadratic formula.

n

_{1,2}={-2±√[(-2)^{2}-4*1*(-8)]}/2n

_{1}=[-2+√36]/2=(-2+6)/2=2n

_{2}=[-2-√36]/2=(-2-6)/2=-4**Answer: n=-4 or n=2**

## 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 !Comment