(x +1) (x+2)( x+3 )(x+4) = 120. Solution using quadratic equation knowledge?

Question

Is the solution of the given problem possible using knowledge of quadratic equations.
How do we find x???

Answer 1

If you wish to use quadratic equations to solve this , note that

(x+1)(x+2)(x+3(x+4) can be transformed into a quadratic

because (x+1)(x+4) = x^2+5x+4, (x+2)(x+3) = x^2+5x+6, so

the entire product is: (x^2+5x+5 +1)(x^2+5x+5 - 1) giving,

(x^2+5x+5)^2 - 1 = 120. Solving this quadratic gives

two more: x^2+5x+5 = +11, x^2+5x+5 = -11. The first one

yields x= +1,- 6 the second yields x= ( - 5 +/- sqrt( - 39) )/2.

Answer 2

1

Answer 3

Rewrite the equation as:

(x + 1)(x + 4)(x + 3)(x + 2) = 120

Multiply the first 2 and last 2 expressions:

(x^2 + 5x + 4)(x^2 + 5x + 6) = 120 ..... (1)

Let x^2 + 5x = y
(1) becomes,

(y + 4)(y + 6) = 120
y^2 + 10x + 24 = 120
y^2 + 10x - 96 = 0
y^2 + 16x - 6y - 96 = 0
y(y + 16) - 6(y + 16) = 0
(y + 16)(y - 6) = 0

y = -16 or 6

x^2 + 5x = 16 or 6

When x^2 + 5x = 6
x^2 + 5x - 6 = 0
x^2 + 6x - x - 6 = 0
x(x + 6) - 1(x + 6) = 0
(x + 6)(x - 1) = 0
x = -6, 1

When x^2 + 5x = -16
x^2 + 5x + 16 = 0

Using quadratic formula:
x = [-5 +/- sqrt (25 - 64)]/2

The solutions are, -6, 1, [-5 - sqrt (25 - 64)]/2 and [-5 + sqrt (25 - 64)]/2
These are all the possible solutions. But the only real solutions are -6 and 1.

Answer 4

Simple
120 = 5! so you know x =1 is one solution , but it wont give you all complex solutions.

Now by quadratic, just group two of them. (in some order)

replace the variable term because they are equal by another
variable, if you don't get try regrouping.

(x+1) (x+2) (x+3) (x+4)
= (x+1) (x+4) . (x+2)(x+3)
= (x^2+5x+ 4) . (x^2+5x+6)
replace m = x^2+5x+4
m*(m+2) = 120

quadratic solution to this is m = 10, -12

now x^2 + 5x + 4 = 10 // -12 is infeasible
or x^2 + 5x - 6 = 0
try with -12 to get more solutions but they will be complex.

quadratic solution to this is x = 1,-6
now you got a negative solution also...

So finally x = 1 or -6

(with -12 is difficult to compute I left, you can try)

Answer 5

(x+1)(x+2)(x+3)(x+4) = 120
beacuse 1 + 4 = 2 + 3 we rearrange

((x+1)(x+4))(x+2)(x+3)) = 120
(x^2+5x +4)(x^2+5x+6) = 120
let x^2+5x+4 = t

t(t+2) = 120
t^2+2t - 120 = 0
(t+12)(t-10) = 0
t = -12 or 10

now t = 10 gives

x^2+5x+4 = 10
x^2+5x - 6 = 0
(x-1)(x+6) = 0 x = 1 or -6
x =1 gives 4 numbers 2 3 4 5 (product 120)
x = -6 gives 4 numbers -5 -4 -3 -2(product 120)
t= 12
gives
x^2+5x+16 = 0

this gives complex roots

Answer 6

yes of course it is possible.
first u should check the constant terms of the equation as in this ques they are 1,2,3,4. now 1+4 = 3+2
so according to this we can rearrange the equation to
(x + 1) (x + 4) (x + 3) (x + 2) = 120
now multiplying first two terms and then last two terms. we get,
(x^2 + 5x + 4) (x^2 + 5x + 6) = 120
putting (x^2 + 5x) = y
the equation becomes
(y + 4) (y + 6) = 120
y^2 + 10y + 24 = 120
find y through this equation and finally find the value of x by substituting the value of y.

Answer 7

(x+1)(x+2)(x+3)(x+4)=120
=(x+1)(x+4)(x+2)(x+3)=120
=(x^2+5x+4)(x^2+5x+6)

taking x^2+5x=y,we have,

(y+4)(y+6)=120
=y^2+10y+24=120
=y^2+10y-96=0
=(y+16)(y-6)=0

y=6 or -16

taking y=6,we have,
x^2+5x=6
=x^2+5x-6=0
=(x+6)(x-1)=0

x=1or -6

taking y= -16,we have,
x^2+5x= -16
=x^2+5x+16=0
this equation has no real solution.

so,x=1 or -6

the equation is a quartic equation(degree 4) so it must have 4 solutions.
But its 2 solutions are real and the other 2 are imaginary.

Answer 8

There is no solution that way because its not a quadratic equation your going to have an "x^4" term. Equations with these high routes don't have simple algorithms to solve them. There might be some complicated ones but I've never seen nor do i want to see it .....then again if someone does know how to solve this using the quadratic formula i would like to know. Maybe they went over it in on of the many classes that I kinda missed

Answer 9

The way this is set up, you're looking for 4 consecutive numbers which multiply to give you 120, which are 2,3,4 and 5. In order for the equation to give you those numbers, x would have to equal 1 (quadratics not needed).

Answer 10

Rearrange the question as (x+1)(x+4)(x+2)(x+3)=120
u see the constants 1and 4, and 2 and 3 both add upto 5
then we get
(x^2+5x+4)(x^2+5x+6)=120
then let m=x^2+5x
so we have
(m+4)(m+6)=120
m^2+10m+24=120
then just substitute the value of m and u will get the answer.