If the roots of the equation \( A x^{2}+B x+C=0 \) are \( \alpha \) an

1.	If the roots of the equation \( A x^{2}+B x+C=0 \) are \( \alpha \) and \( \beta \), Show that the equiation whose roots are \( \frac{\alpha}{\beta} \) and \( \frac{\beta}{\alpha} \) is given by\[A C x^{2}-\left(B^{2}-2 A C\right) x+A C=0\]
Answer» Roots of the equation Ax² + Bx + C = 0 are α and β. ∴ Sum of roots = α + β = \(\frac{-B}{A}\) And product of roots = αβ = \(\frac{C}{A}\) Now, (α + β)² = α² + β² + 2αβ ∴ α² + β² = (α + β)²- 2αβ = \((\frac{-B}{A})^2\)- \(\frac{2C}{A}\) = \(\frac{B^2}{A^2}\) - \(\frac{2C}{A}\) = \(\frac{B^2-2AC}{A^2}\) Sum of required roots \(\frac{\alpha}{\beta}\) and \(\frac{\beta}{\alpha}\), = \(\frac{\alpha}{\beta}\) + \(\frac{\beta}{\alpha}\) = \(\frac{\alpha^2+\beta^2}{\alpha\beta}\) = \(\frac{\frac{B^2-2AC}{A^2}}{\frac{C}{A}}\) = \(\frac{B^2-2AC}{AC}\) And products of roots = \(\frac{\alpha}{\beta}\) x \(\frac{\beta}{\alpha}\) = 1. ∴ Equation whose roots we are \(\frac{\alpha}{\beta}\) and \(\frac{\beta}{\alpha}\) is : x² - (sum of roots)x + Product of roots = 0 ⇒ x² - \(\frac{(B^2-2AC)}{AC}x\) + 1 = 0 ⇒ ACx² - (B²-2AC)x + AC = 0 Hence proved.

If the roots of the equation \( A x^{2}+B x+C=0 \) are \( \alpha \) and \( \beta \), Show that the equiation whose roots are \( \frac{\alpha}{\beta} \) and \( \frac{\beta}{\alpha} \) is given by\[A C x^{2}-\left(B^{2}-2 A C\right) x+A C=0\]

Answer»

Roots of the equation Ax² + Bx + C = 0 are α and β.

∴ Sum of roots = α + β = \(\frac{-B}{A}\)

And product of roots = αβ = \(\frac{C}{A}\)

Now,

(α + β)² = α² + β² + 2αβ

∴ α² + β² = (α + β)²- 2αβ

= \((\frac{-B}{A})^2\)- \(\frac{2C}{A}\)

= \(\frac{B^2}{A^2}\) - \(\frac{2C}{A}\)

= \(\frac{B^2-2AC}{A^2}\)

Sum of required roots \(\frac{\alpha}{\beta}\) and \(\frac{\beta}{\alpha}\),

= \(\frac{\alpha}{\beta}\) + \(\frac{\beta}{\alpha}\)

= \(\frac{\alpha^2+\beta^2}{\alpha\beta}\)

= \(\frac{\frac{B^2-2AC}{A^2}}{\frac{C}{A}}\)

= \(\frac{B^2-2AC}{AC}\)

And products of roots = \(\frac{\alpha}{\beta}\) x \(\frac{\beta}{\alpha}\) = 1.

∴ Equation whose roots we are \(\frac{\alpha}{\beta}\) and \(\frac{\beta}{\alpha}\) is :

x² - (sum of roots)x + Product of roots = 0

⇒ x² - \(\frac{(B^2-2AC)}{AC}x\) + 1 = 0

⇒ ACx² - (B²-2AC)x + AC = 0

Hence proved.