Question

If α and β are the roots of the equation 3x²-6x+1=0 from the equation whose

roots are α²β, β²α.​

Answer 1

Answer:

• Gn
• $3x^{2} - 6x + 1 = 0$
• Therefore
• $\alpha = 6 \div 3 = 2$
• $\beta = 1 \div 3$
• Now apply the value
• $\alpha {}^{2} \beta = 4 \times 1 \div 3 = 4 \div 3$
• $\beta {}^{2} \alpha = < 1 \div 3 > {}^{2} \times 2 \\ = 2 \div 9$

Answer 2

Given :

• α and β are the roots of the equation 3x²-6x+1=0

To find :

• equation whose roots are α²β and β²α

Concept and formula used :

$\underline{\boxed{\bf{ sum \: of \: roots = - \frac{coefficient \: of \: {x}}{coefficient \: of \: {x}^{2} } }}}$

$\underline{\boxed{\bf{product \: of \: roots = \frac{constant \: term}{coefficient \: of \: {x}^{2} } }}}$

$\sf \: general \: equation \: \\ \sf \: can \: be \: written \: of \: the \: \sf \: form \: : \\ \sf{x}^{2} - (sum \: of \: roots)x + (product \: of \: roots)$

Solution :

We have :-

$\sf \: 3 {x}^{2} - 6 + 1 = 0$

α and β are roots of above equation.

$\sf{ sum \: of \: roots = \alpha + \beta = - (\dfrac{ - 6}{3 })}$

$\sf{ sum \: of \: roots = \alpha + \beta = 2}$

${ \sf{product \: of \: roots = \alpha \beta = \dfrac{1}{3} }}$

$\sf \: Now \: if \: { \alpha}^{2} \beta \: and \: { \beta}^{2} \alpha \: are \: roots \: \\ \sf then \: equation \: will \: be \: of \: form :$

$\implies \sf{x}^{2} - ({ \alpha}^{2} \beta + \alpha{ \beta}^{2})x + ({ \alpha}^{2} \beta \times \alpha { \beta}^{2}) = 0$

$\implies \sf{x}^{2} - \alpha\beta({ \alpha} + { \beta})x + ({ \alpha}^{3} { \beta}^{3}) = 0$

$\implies \sf{x}^{2} - \alpha\beta({ \alpha} + { \beta})x + {(\beta \alpha)}^{3} = 0$

Now put :-

$\alpha + \beta = 2$

$\alpha \beta = \dfrac{1}{3}$

$\implies \sf{x}^{2} - \dfrac{2}{3} x + {( \dfrac{1}{3} )}^{3} = 0$

$\implies \sf{x}^{2} - \dfrac{2}{3} x + {\dfrac{1}{27} } = 0$

LCM if 27 and 3 = 27

$\implies \dfrac{ \sf27{x}^{2} - 18x + 1}{27} = 0$

$\implies { \sf27{x}^{2} - 18x + 1} = 0$

Answer :

${ \sf27{x}^{2} - 18x + 1} = 0$