If α, β are the roots of the quadratic equation x2+ax+b=0, (b≠ 0); then the quadratic equation whose roots are α -(1/β), β - (1/α) is

Question

If α, β are the roots of the quadratic equation x2+ax+b=0, (b≠ 0); then the quadratic equation whose roots are α -(1/β), β - (1/α) is (A) ax2+a(b-1)x+(a-1)2=0 (B) bx2+a(b-1)x+(b-1)2=0 (C) x2+ax+b = 0 (D) abx2+bx+a = 0

Tardigrade Admin · Accepted Answer

Given equation is, x2+a x+b=0,(b ≠ 0) its roots are α and β. Then, sum of roots =α+β=-a ....(i) Product of roots =α ⋅ β=b .....(ii) Now, (α-(1/β))+(β-(1/α))=(α+β)-((α+β/α β)) =-a-((-a)/b) [from Eqs.(i) and (ii)] =-a+(a/b)=(a/b)(1-b) and (α-(1/β))(β-(1/α))=α β-1-1+(1/α β) =b+(1/b)-2[ text from Eq. (ii) ] ....(iv) =(1/b)(b2-2 b+1)=(1/b)(b-1)2 ∴ Required of quadratic equation whose roots are (α-(1/β)) and (β-(1/α)) is x2- (α-(1/β))+(β-(1/α)) x + (α-(1/β))(β-(1/α)) =0 On putting the values from Eqs. (i) and (ii), we get x2-(a/b)(1-b) x+(1/b)(b-1)2=0 ⇒ b x2+a(b-1) x+(b-1)2=0, b ≠ 0

Q. If $α, β$ are the roots of the quadratic equation $x^{2} + a x + b = 0, (b \neq = 0)$ ; then the quadratic equation whose roots are $α - \frac{1}{β}, β - \frac{1}{α}$ is

$a x^{2} + a (b - 1) x + (a - 1)^{2} = 0$

$b x^{2} + a (b - 1) x + (b - 1)^{2} = 0$

$x^{2} + a x + b = 0$

$ab x^{2} + b x + a = 0$

Q. If α,β are the roots of the quadratic equation x2+ax+b=0,(b=0); then the quadratic equation whose roots are α−β1​,β−α1​ is

ax2+a(b−1)x+(a−1)2=0

bx2+a(b−1)x+(b−1)2=0

x2+ax+b=0

abx2+bx+a=0

Q. If $α, β$ are the roots of the quadratic equation $x^{2} + a x + b = 0, (b \neq = 0)$ ; then the quadratic equation whose roots are $α - \frac{1}{β}, β - \frac{1}{α}$ is

$a x^{2} + a (b - 1) x + (a - 1)^{2} = 0$

$b x^{2} + a (b - 1) x + (b - 1)^{2} = 0$

$x^{2} + a x + b = 0$

$ab x^{2} + b x + a = 0$