If sum of values of ′x′ of equation x23+x13−2=0 is α . Then find |α|.
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Solution
x23+x13−2=0
In this case we can see that,
23=2(13)
and so one of the exponents is twice the other so it looks like we’ve got an equation that is reducible to quadratic in form. The substitution will then be
u=x13
∴u2=x23
Substituting this into the equation gives,
u2+u−2=0
By using formula −b±√b2−4ac2a we get factors,
u=−1+√3 and u=−1−√3
So, x13=−1+√3
∴x=(−1+√3)3=(−1)3+(√3)3+3(−1)(√3)(−1+√3)
∴x=−1+3√3−3√3(−1+√3)=−10+6√3 ----- ( 1 )
⇒ Now, x13=(−1−√3)
x=(−1−√3)3=(−1)3−(√3)3−3(−1)(√3)(−1−√3)
∴x=−10−6√3 ------ ( 2 )
⇒ Sum of values of x=(−10+6√3)+(−10−6√3) [ From ( 1 ) and ( 2 ) ]