Answer to Question #21736 in Algebra for arifa

Question

(x-2)(x-4)(x-6)(x-8)=9

Answer 1

(x-2)(x-4)(x-6)(x-8)=9
[ (x-2)(x-8)] *[ (x-4)(x-6)]=9
(x^2-10x+16)*(x^2-10x+24)=9
let t=x^2-10x+16
t(t+8)=9
t1=1, t2=-9
so we have
x^2-10x+16=1 orx^2-10x+16=-9
x^2-10x+15=0 or x^2-10x+25=0
second x^2-10x+25=0 so (x-5)^2=0 x=5
x^2-10x+15=0 D=100-60=40 x1,2=(10+-2sqrt(10))/2=5+-sqrt(10)
we have three different roots
x1=5+sqrt(10)
x2=5-sqrt(10)
x3=5