2n^2+4n-16=0

Question

Solving quadratic functions with equations

Answer 1

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)

Answer 2

Hi Raven;

2n²+4n-16=0

For the FOIL...

FIRST must be (2x)(n)=2n².

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)=2n²

OUTER...(2n)(-2)=-4n

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

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

2n²-4n+8n-16=0

2n²+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

 

Answer 3

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

 

n²+2n-8=0

 

Now 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)]}/2

n₁=[-2+√36]/2=(-2+6)/2=2

n₂=[-2-√36]/2=(-2-6)/2=-4

 

Answer: n=-4 or n=2