Question #21736

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

Expert's answer

(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

[ (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

## Comments

Assignment Expert31.01.13, 13:50You're welcome. We are glad to be helpful.

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arifa31.01.13, 07:07thanks for such a nice help.

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