Question
6(x2+1x2)−25(x−1x)+12=0
Answer: Option A
:
A
Let x−1x=z
(x−1x)2=z2
(x2+1x2−2)=z2
x2+1x2=z2+2
Now substituting in the equation:
6(z2+2)−25z+12=0
6z2+12−25z+12=0
6z2−25z+24=0
6z2−16z−9z+24=0
2z(3z−8)−3(3z−8)=0
(2z−3)(3z−8)=0
z=32;z=83
x−1x=32
x−1x=83
2x2=3x−2=0
3x2−8x−3=0
2x2−4x+x−2=0
3x2−9x+x−3=0
2x(x−2)+1(x−2)=0
3x(x−3)+1(x−3)=0
(2x+1)(x−2)=0
(3x+1)(x−3)=0
x=−12,x=2
x=−13,x=3
So,the roots are:−12,2,−13,3
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:
A
Let x−1x=z
(x−1x)2=z2
(x2+1x2−2)=z2
x2+1x2=z2+2
Now substituting in the equation:
6(z2+2)−25z+12=0
6z2+12−25z+12=0
6z2−25z+24=0
6z2−16z−9z+24=0
2z(3z−8)−3(3z−8)=0
(2z−3)(3z−8)=0
z=32;z=83
x−1x=32
x−1x=83
2x2=3x−2=0
3x2−8x−3=0
2x2−4x+x−2=0
3x2−9x+x−3=0
2x(x−2)+1(x−2)=0
3x(x−3)+1(x−3)=0
(2x+1)(x−2)=0
(3x+1)(x−3)=0
x=−12,x=2
x=−13,x=3
So,the roots are:−12,2,−13,3
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