Search 75,768 tutors
FIND TUTORS
Ask a question
0 0

2n^2+4n-16=0

Tutors, please sign in to answer this question.

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;
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
 
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.
n1,2={-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
 

Comments

Hi, Raven:
 
What you wrote is called quadratic equation, not quadratic formula. The second thing, you shall think generally, not in specifics. If I write yx2+zx+s=0, this is still quadratic equation. And m*sin2(x)+h*sin(x)+w=0 is quadratic equation, with respect to sin(x), of course.
Look at what Kirill did:divided both sides of the equation by 2 to get n^2+2n-8=0.
Another way to look at the problem now is that we want two numbers whose product is 8(disregarding signs for the moment). Because 2 is even the two factors for 8 must both be even which eliminates 1 and 8 and leaves us with only 2 and 4.
4*(-2)=-8 and 4+(-2)=2
Always factor out any numerical common factors and make sure it's the GCF. This will make your work simpler. Always put your polynomial (equation) in the form ax^2+bx+c(=0).
Look at a,b, and c. Each one is either even(E) or odd(O). There are 8 possibilities for a,b, and c:
OOO
EEE
EEO
OEE
EOE
OOE
EOO
OEO
There are rules for 7 of the 8 possibilities.
I gave you one already.
Another:if you have OOE, you must factor E into an odd times an even number; you can't use two even numbers ! This eliminates some of the possibilities and makes your work easier.
Another example:If you have EEO, you must factor the first E using 2 even numbers, not an even times an odd !
If you have OOO and the three coefficients have no common factors, then the polynomial is
irreducible (and prime) over the set of integers. You must then use the quadratic formula.
Try to figure out the other rules. By the way, there is no rule for OEO !

Comment