Question

# Solve the equations:3x4−10x3+4x2−x−6=0, one root being 1+√−32.

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Solution

## p(x)=3x4−10x3+4x2−x−6 One of the root is 1+√−32 then one other root is 1−√−32 as imaginary roots occur in pair ⇒(x−1+√−32)(x−1−√−32) is a factor of p(x) ⇒(x2−x+1) is a factor of p(x) Dividing p(x) by (x2−x−1) ⇒p(x)=(3x2−7x−6)(x2−x+1)⇒(3x2−7x−6)(x2−x+1)=0⇒3x2−7x−6=0⇒3x2−9x+2x−6=0⇒3x(x−3)−2(x−3)=0⇒(3x−2)(x−3)=0⇒x=3,32 So the other two roots are 3,32

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