# Demonstrate if ( 1 - x1 )( 1 - x2 )( 1 - x3 )( 1 - x4 ) = 5 where x1,2,3,4 roots of the equation .

hala718 | Certified Educator

Since we have 4 roots, then the function is a 4th degree function:

ax^4+bx^3+cx^2+dx+e=0

We know that:

x1 +x2+x3+x4= -b/a=1........(1)

also

x1*x2 + x1*x3+x1*x4+x2*x3+x2*x4+x3*x4=  c/a=1....(2)

and:

x1*x2*x3 + x1*x2*x4 + x1*x3*x4 + x2*x3*x4 = -d/a=1..(3)

and x1*x2*x3*x4 = e/a=1...(4)

Now let us open the brackets:

(1-x1)(1-x2)(1-x3)(1-x4)= (1-x1-x2 +x1*x2)(1-x3-x4+x3*x4)

= 1-x3-x4+x3*x4 -x1+x1*x3+ x1*x4 -x1*x3*x4 -x2+x2*x3 +x2*x4 -x2*x3*x4 +x1*x2 -x1*x2*x3 -x1*x2*x4 +x1*x2*x3 +x1*x2*x3*x4

Now we will substitute with (1) (2) (3) and (4):

= 1+ b/a + c/a+d/a +e/a = 5

neela | Student

Let us supoose that  the required 4th degree equation is x^4+ax^3+bx^2+cx+d, whose roots are x1,x2,x3 and x4,

Then sum of roots  =-a

sum of product of roots taken 2 at a time = b

sum of the product of roots taken 3 at a time = -c

prduct of roots = d

Now (1-x1)((1-x2)(1-x3)(1-x4) = d. Expanding the LHS, we get:

1-(sum of the roots)+(product of roots taken 2 at a time)- product of roots taken 3 at a time) + oroduct of all 4 roots = 5. Or

1-(-a)+b-(-c)+d = 5. Or

1+a+b+c+d = 5. Or

a+b+c+d = 4 is the condition.

Example for the demonstration:

(x-2)^3 *(x-6) = o has roots x1=x2=x3= 2 and x4 = 6

Then (x-2)^3(x-6) = x^4 -12x^3+48x^2-80x+48 = 0.

a= -12,b =48,c =-80 and d = 48.

(a+b+c+d) = -12+48-80+48 = 4.

(1-x1)(1-x2)(1-x3)(1-x4) = (1-2)(1-2)(1-2)(1-6) = (-1)(-1)(-1)(-5) = 5.

So the there exists the 4th degree equation, x^4+ax^3+bx^3+cx+d = 0,whosex1,x2,x3 and x4 can satisfy (1-x1)(1-x2)(1-x3)(1-x4) = 5, if a+b+c+d = 4.

giorgiana1976 | Student

Since we have 4 roots, that means that the equation is a fourth degree polynomial ax^4 + bx^3 + cx^2 + dx + e = 0.

Opening the brackets, we'll obtain the polynomial  and after that we could write  Viete's relations:

x1 + x2 + x3 + x4 = -b/a=-1

x1*x2 + x1*x3 + x1*x4 + x2*x3 + x2*x4 + x3*x4=c/a=1

x1*x2*x3 + x1*x2*x4 + x1*x3*x4 + x2*x3*x4 = -d/a=1

x1*x2*x3*x4= e/a=1

(1-x1)(1-x2)(1-x3)(1-x4)=1-(x1 + x2 + x3 + x4) + (x1*x2 + x1*x3 + x1*x4 + x2*x3 + x2*x4 + x3*x4) - (x1*x2*x3 + x1*x2*x4 + x1*x3*x4 + x2*x3*x4) + (x1*x2*x3*x4)=

= 1+b/a+c/a+d/a+e/a=1+1+1+1+1=5 true