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Mathematics Question on Complex Numbers and Quadratic Equations

If α,β\alpha, \beta are the roots of the quadratic equation x2+ax+b=0,(b≠0)x^2+ax+b=0, (b\ne 0); then the quadratic equation whose roots are α−1β,β−1α\alpha -\frac{1}{\beta}, \beta - \frac{1}{\alpha} is

A

ax2+a(b−1)x+(a−1)2=0ax^2+a(b-1)x+(a-1)^2=0

B

bx2+a(b−1)x+(b−1)2=0bx^2+a(b-1)x+(b-1)^2=0

C

x2+ax+b=0x^2+ax+b = 0

D

abx2+bx+a=0abx^2+bx+a = 0

Answer

bx2+a(b−1)x+(b−1)2=0bx^2+a(b-1)x+(b-1)^2=0

Explanation

Solution

The correct option is(B): bx 2+a(b−1)x+(b−1)2=0.

Given equation is, x2+ax+b=0,(b≠0)x^{2}+a x+b=0,(b \neq 0)
its roots are α\alpha and β\beta.
Then, sum of roots =α+β=−a=\alpha+\beta=-a ....(i)
Product of roots =α⋅β=b=\alpha \cdot \beta=b .....(ii)
Now,
(α−1β)+(β−1α)=(α+β)−(α+βαβ)\left(\alpha-\frac{1}{\beta}\right)+\left(\beta-\frac{1}{\alpha}\right)=(\alpha+\beta)-\left(\frac{\alpha+\beta}{\alpha \beta}\right)
=−a−(−a)b=-a-\frac{(-a)}{b} [from Eqs.(i) and (ii)]
=−a+ab=ab(1−b)=-a+\frac{a}{b}=\frac{a}{b}(1-b)
and (α−1β)(β−1α)=αβ−1−1+1αβ\left(\alpha-\frac{1}{\beta}\right)\left(\beta-\frac{1}{\alpha}\right)=\alpha \beta-1-1+\frac{1}{\alpha \beta}
=b+1b−2[ from E (ii) ]=b+\frac{1}{b}-2[\text { from E (ii) }] ....(iv)
=1b(b2−2b+1)=1b(b−1)2=\frac{1}{b}\left(b^{2}-2 b+1\right)=\frac{1}{b}(b-1)^{2}
∴\therefore Required of quadratic equation whose roots are (α−1β)\left(\alpha-\frac{1}{\beta}\right) and (β−1α)\left(\beta-\frac{1}{\alpha}\right) is
x^{2}-\left\\{\left(\alpha-\frac{1}{\beta}\right)+\left(\beta-\frac{1}{\alpha}\right)\right\\} x
+\left\\{\left(\alpha-\frac{1}{\beta}\right)\left(\beta-\frac{1}{\alpha}\right)\right\\}=0
On putting the values from Eqs. (i) and (ii), we get
x2−ab(1−b)x+1b(b−1)2=0x^{2}-\frac{a}{b}(1-b) x+\frac{1}{b}(b-1)^{2}=0
⇒bx2+a(b−1)x+(b−1)2=0,b≠0\Rightarrow \quad b x^{2}+a(b-1) x+(b-1)^{2}=0, b \neq 0