Let a and b are non-zero real numbers and α3+β3=- a , α β= b, then the quadratic equation whose roots are (α2/β), (β2/α) is

Question

Let a and b are non-zero real numbers and α3+β3=- a , α β= b, then the quadratic equation whose roots are (α2/β), (β2/α) is (A) a x2+b x+a2=0 (B) b x2-a x+b2=0 (C) b x2+a x+b2=0 (D) a x2-b x+a2=0

Tardigrade Admin · Accepted Answer

α3+β3=-a, α β=b Now, (α2/β)+(β2/α)=(- a / b ) ; α β= b Equation of x2-((-a/b)) x+b=0 ⇒ b x2+4 x+b2=0

Q. Let $a$ and $b$ are non-zero real numbers and $α^{3} + β^{3} = - a, α β = b$ , then the quadratic equation whose roots are $\frac{α ^{2}}{β}, \frac{β ^{2}}{α}$ is

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

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

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

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

$α^{3} + β^{3} = - a, α β = b$
Now, $\frac{α ^{2}}{β} + \frac{β ^{2}}{α} = \frac{- a}{b}; α β = b$
Equation of $x^{2} - (\frac{- a}{b}) x + b = 0 \Rightarrow b x^{2} + 4 x + b^{2} = 0$

Q. Let a and b are non-zero real numbers and α3+β3=−a,αβ=b, then the quadratic equation whose roots are βα2​,αβ2​ is

ax2+bx+a2=0

bx2−ax+b2=0

bx2+ax+b2=0

ax2−bx+a2=0