If (1/√α) and (1/√β) are the roots of the equation, ax2 + bx +1 = 0 (a ≠ 0, a, b ∈ R), then the equation, x(x + b3) + (a3 â 3abx) = 0 has roots :

Question

If (1/√α) and (1/√β) are the roots of the equation, ax2 + bx +1 = 0 (a ≠ 0, a, b ∈ R), then the equation, x(x + b3) + (a3 â 3abx) = 0 has roots : (A) α3/2 and β3/2 (B) α β1/2 and α1/2 β (C) √α β and α β (D) α-(3/2) and β-(3/2)

Tardigrade Admin · Accepted Answer

(1/√α)+(1/√β)=-(b/a) also (1/√αβ )=(1/a) ⇒ √α+√β=-b now x (x + b3) + a3 - 3abx =x2 + (b3 - 3ab) x + a3 = x2+ b (b2 - 3a) x +a3 =x2-(√α +√β ) α+β+2√αβ -3√α β x+αβ √αβ =x2-(α√α +β√ β )+αβ√α β ⇒ roots are α √α and β √ β

Q. If $\frac{1}{α}$ and $\frac{1}{β}$ are the roots of the equation, $a x^{2} + b x + 1 = 0 (a \neq = 0, a, b \in R)$ , then the equation, $x (x + b^{3}) + (a^{3} —3 ab x)$ = 0 has roots :

$α^{3/2}$ and $β^{3/2}$

$α β^{1/2}$ and $α^{1/2} β$

$α β$ and $α β$

$α^{- \frac{3}{2}}$ and $β^{- \frac{3}{2}}$

Q. If α​1​ and β​1​ are the roots of the equation, ax2+bx+1=0(a=0,a,b∈R), then the equation, x(x+b3)+(a3—3abx) = 0 has roots :

α3/2 and β3/2

αβ1/2 and α1/2β

αβ​ and αβ

α−23​ and β−23​

α​1​+β​1​=−ab​alsoαβ​1​=a1​ ⇒α​+β​=−b now x(x+b3)+a3−3abx =x2+(b3−3ab)x+a3 =x2+b(b2−3a)x+a3 =x2−(α​+β​){α+β+2αβ​−3αβ​}x+αβαβ​ =x2−(αα​+ββ​)+αβαβ​ ⇒ roots are αα​ and ββ​

Q. If $\frac{1}{α}$ and $\frac{1}{β}$ are the roots of the equation, $a x^{2} + b x + 1 = 0 (a \neq = 0, a, b \in R)$ , then the equation, $x (x + b^{3}) + (a^{3} —3 ab x)$ = 0 has roots :

$α^{3/2}$ and $β^{3/2}$

$α β^{1/2}$ and $α^{1/2} β$

$α β$ and $α β$

$α^{- \frac{3}{2}}$ and $β^{- \frac{3}{2}}$