If a pm ib and b pm ai are the roots of x^4-1 | vedclass

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(C) The given equation is $x^4-10x^3+50x^2-130x+169=0$. Since the coefficients are real,the roots occur in conjugate pairs. The roots are $a+ib, a-ib,

Vedclass · Accepted Answer

$\frac{13}{6}$

Vedclass · Answer

(C) The given equation is $x^4-10x^3+50x^2-130x+169=0$. Since the coefficients are real,the roots occur in conjugate pairs. The roots are $a+ib, a-ib, b+ai, b-ai$. The sum of the roots is $(a+ib) + (a-ib) + (b+ai) + (b-ai) = 2a + 2b = 10$,so $a+b=5$. The product of the roots is $(a^2+b^2)(b^2+a^2) = (a^2+b^2)^2 = 169$. Thus,$a^2+b^2 = 13$. We know that $(a+b)^2 = a^2+b^2+2ab$. Substituting the values,$5^2 = 13 + 2ab$,which gives $25 = 13 + 2ab$,so $2ab = 12$ or $ab = 6$. We need to find $\frac{a}{b} + \frac{b}{a} = \frac{a^2+b^2}{ab}$. Substituting the values,$\frac{13}{6}$.

If $a \pm ib$ and $b \pm ai$ are the roots of $x^4-10x^3+50x^2-130x+169=0$,then $\frac{a}{b}+\frac{b}{a}=$

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