Solve (x+1)(x+2)(x+3)(x+4)+1 = 0

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01 May 2014 Quadratic Equation

Solve
(x+1)(x+2)(x+3)(x+4)+1 = 0

OPtion

1) {-5+sqrt(5)}/2, {-5-sqrt(5)}/2
2) {-2+sqrt(2)}/5, {-2-sqrt(2)}/5
3) {-4+sqrt(2)}/2, {-4-sqrt(2)}/2
4) {-6+sqrt(2)}/4, {-6-sqrt(2)}/4
5) {-4+sqrt(5)}/2, {-4-sqrt(5)}/2
6) {-3+sqrt(3)}/2, {-3-sqrt(3)}/2
7) {-3+sqrt(5)}/2, {-3-sqrt(5)}/2
8) {-5+sqrt(5)}/4, {-5-sqrt(5)}/4
9) {-2+sqrt(2)}/3, {-2-sqrt(2)}/3
10)None of these

Solution

Multiply first factor with last and second with third factor to get
(x^2+5x+4)(x^2+5x+6)+1 = 0
Let x^2+5x = t
then
(t+4)(t+6)+1 = 0
Solve
(t+5)^2 = 0
t = -5
x^2+5x = -5
x^2+5x+5 = 0
Solve
Option 1)

Correct Option: 1 Best Solution (16)

ABHOY KUMAR   10 years AGO

(x+1)(x+2)(x+3)(x+4)+1=0
(x+1)(x+4)(x+2)(x+3)+1=0
(x^2+5x+4)(x^2+5x+6)+1=0
let x^2+5x=t ......(i)
(t+4)(t+6)+1=0
t^2+10t+24+1=0
t^2+10t+25=0
(t+5)^2=0
t=-5
put t=-5 in eq.(i)
x^2+5x=-5
x^2+5x+5=0
x=(-5+sqrt(5))/2
and x=(-5-sqrt(5))/2
{-5+sqrt(5)}/2, {-5-sqrt(5)}/2[answer]

Like? Yes (16) | No (7)  

Ann Theressa   10 years AGO

(x+1)(x+2)(x+3)(x+4)+1 = 0
(x^2+3x+2)(x^2+7x+12) + 1 = 0
x^4 + 10x^3 + 35x^2 + 50x + 25 = 0
(x^2 + 5x + 5)^2 = 0

Taking square root on both sides

x^2 + 5x + 5 = 0
x = (-5 (+-) sqrt(25-20)}/2

x = {-5-sqrt(5)}/2 ,{-5+sqrt(5)}/2

Like? Yes (22) | No (15)  

PAWAN KUMAR   10 years AGO

solving using options,
1st option:
(-4*4)/16 +1=0
here, LHS=RHS
hence answer 1 is correct.

Like? Yes (5) | No (1)  

llamar   10 years AGO

by trail and error method
the given equation is clearly
(x+1)(x+2)(x+3)(x+4)+1 =(x^2+5x+5)^2=0
x^2+5x+5=0 by solving it we get
{-5+sqrt(5)}/2, {-5-sqrt(5)}/2 as roots

Like? Yes (6) | No (2)  

vishnuvardhan   10 years AGO

(x+1)(x+4)(x+2)(x+3)+1=0
(x^2+5x+4)*(x^2+5x+6)+1=0
put x^2+5x=t
(t+4)*(t+6)+1=0
t=-5
x^2+5x=-5
x=(-5+sqrt5)/2,(-5-sqrt5)/2

Like? Yes (5) | No (2)  

SIBNATH HALDER   10 years AGO

{(x+1)(x+4)}{(x+2)(x+3)}+1=0
we get
(x^2+5x+4)(x^2+5x+6)+1=0
let x^2+5x=a
(a^2+4)(a^2+6)+1=0
a=+5,-5;
put value of a=-5
x^2+5x+5=0
x=option(1)
ans

Like? Yes (5) | No (3)  

Dr.Dipin Singh   10 years AGO

(x+1)(x+2)(x+3)(x+4)+1 = 0
(x+1)(x+4)(x+2)(x+3)+1 = 0
(x^2+5x+4)*(x^2+5x+6)+1=0
Let x^2+5x=y
(y+4)*(y+6)+1=0
y^2+10+25=0
(y+5)^2=0
y=-5
x^2+5x+5=0
solving ,
we get
x= {-5+sqrt(5)}/2, {-5-sqrt(5)}/2... option 1

Like? Yes (6) | No (5)  

azharshahid01@gmail.com   10 years AGO

On putting values of x in the equation.
only option(1) will satisfy the equation.

Like? Yes (7) | No (6)  

VIMALRAJ V   10 years AGO

on simplification x⁴ + 10x³ + 35x² + 50x + 25 = 0.
then by using long division method, the quotient becomes x^2 + 5x + 5 = 0.
then by factorization method the answer obtained is {-5+sqrt(5)}/2, {-5-sqrt(5)}/2.
therefore option 1 is correct....

Like? Yes (3) | No (2)  

nandini keshari   10 years AGO

(x+1)(x+2)(x+3)(x+4)+1=0
(x+1)(x+4)(x+2)(x+3)+1=0
(x^2+5x+4)(x^2+5x+6)+1=0
Let x^2+5x=p
(P+4)(P+6)+1=0
P^2+10P+25=0
(P+5)^2=0
P+5=0
x^2+5x+5=0
So
x=-5-sqrt(5)/2 or -5+sqrt(5)/2

Like? Yes (4) | No (3)  

Purosotam Gupta   10 years AGO

put the value of x in equation it satisfied.

Like? Yes (3) | No (3)  

RAKESH   10 years AGO

(x+1)(x+2)(x+3)(x+4)+1 = 0
=> (x^2+5x+4)(x^2+5x+6)+1=0
=> (y+4)(y+6)+1=0
=> y^2+10y+25=0
=> y+5=0
=> x^2+5x+5=0
=> x=[-5+-sqrt(5)]/2

Like? Yes (3) | No (4)  

indrajeet yadav   10 years AGO

(x+1)(x+2)(x+3)(x+4)+1=0
(x+1)(x+4)(x+2)(x+3)=-1
(x^2+5x+4)(x^2+5x+6)=-1
x^2+5x+4=-1 or x^2+5x+6=-1
x^2+5x+5=0 or x^2+5x+7=0
x1={-5+sqrt(5)}/2 or x1={-5+i sqrt(3)}/2
x2={-5-sqrt(5)}/2 or x2={-5-i sqrt(3)}/2
but given option x can't be imaginary.
so x1={-5+sqrt(5)}/2
x2={-5-sqrt(5)}/2

Like? Yes (6) | No (7)  

Tanu   10 years AGO

(x+1)(x+4)(x+2)(x+3)+1=0
(x^2+5x+4)(x^2+5x+6)+1=0--------------------eq1

Let x^2+5x=y--------------------------------eq2

Putting above
(y+4)(y+6)+1=0
y+10y+25=0
y=5

x^2+5x=y
x^2+5x=5
x^2+5x+5=0

x= {-5+sqrt(5)}/2, {-5-sqrt(5)}/2

Like? Yes (2) | No (3)  

MD SOHAIL AKHTER   10 years AGO

on substituting the above option in given eqn, we get;
x= -5+sqrt(5)}/2, {-5-sqrt(5)}/2

so option (1) is correct

Like? Yes (1) | No (4)  

ASHISH MITTAL   10 years AGO

by solving equation for x

Like? Yes (2) | No (5)