Question

If $ \alpha $ , $ \beta $ are the roots of $ a{x^2} + bx + c = 0 $ , then find the values of

I. $ \dfrac{1}{{{\alpha ^2}}} + \dfrac{1}{{{\beta ^2}}} $

II. $ {\alpha ^4}{\beta ^7} + {\alpha ^7}{\beta ^4} $

III. $ {\left( {\dfrac{\alpha }{\beta } - \dfrac{\beta }{\alpha }} \right)^2} $

Answer 1

Hint: For solving this particular question , we consider that for a quadratic equation $ a{x^2} + bx + c = 0 $ , the sum of its roots $ = \dfrac{{ - b}}{a} $ and the product of its roots $ = \dfrac{c}{a} $ Therefore\[\alpha + \beta = \dfrac{{ - b}}{a}\] and $ \alpha \beta = \dfrac{c}{a} $ , with this concept we can solve this question.

Complete step by step solution:

It is given that $ \alpha $ , $ \beta $ are the roots of $ a{x^2} + bx + c = 0 $ , For $ a{x^2} + bx + c = 0 $ , the sum of its roots $ = \dfrac{{ - b}}{a} $ and the product of its roots $ = \dfrac{c}{a} $ . Therefore, \[\alpha + \beta = \dfrac{{ - b}}{a}\] and $ \alpha \beta = \dfrac{c}{a} $ .

Now, we have to find the value of the given equation that is $ \dfrac{1}{{{\alpha ^2}}} + \dfrac{1}{{{\beta ^2}}} $ .

For a quadratic equation $ a{x^2} + bx + c = 0 $ , the sum of its roots $ = \dfrac{{ - b}}{a} $ and the product of its roots $ = \dfrac{c}{a} $ Therefore, \[\alpha + \beta = \dfrac{{ - b}}{a}\] and $ \alpha \beta = \dfrac{c}{a} $ .

Taking the given equation

 $ = \dfrac{1}{{{\alpha ^2}}} + \dfrac{1}{{{\beta ^2}}} = \dfrac{{{\alpha ^2} + {\beta ^2}}}{{{{(\alpha \beta )}^2}}} $

$= \dfrac{{{{(\alpha + \beta )}^2} - 2\alpha \beta }}{{{{(\alpha \beta )}^2}}} $ - (1)

$ = \dfrac{{\dfrac{{{b^2}}}{{{a^2}}} - \dfrac{{2c}}{a}}}{{\dfrac{{{c^2}}}{{{a^2}}}}} = \dfrac{{{b^2} - 2ac}}{{{c^2}}} $

The value of $ \dfrac{1}{{{\alpha ^2}}} + \dfrac{1}{{{\beta ^2}}}  = \dfrac{{{b^2} - 2ac}}{{{c^2}}} $.

II. Now, we have to find the value of the given equation that is $ {\alpha ^4}{\beta ^7} + {\alpha ^7}{\beta ^4} $ .

For a quadratic equation $ a{x^2} + bx + c = 0 $ , the sum of its roots $ = \dfrac{{ - b}}{a} $ and the product of its roots $ = \dfrac{c}{a} $ 

Therefore, \[\alpha + \beta = \dfrac{{ - b}}{a}\] and $ \alpha \beta = \dfrac{c}{a} $ .

Taking the given equation

$  = {\alpha ^4}{\beta ^7} + {\alpha ^7}{\beta ^4} $

$  = {(\alpha \beta )^4}({\alpha ^3} + {\beta ^3}) $

$  = {(\alpha \beta )^4}(\alpha + \beta )({\alpha ^2} + {\beta ^2} - \alpha \beta ) $

We can take the value from the equation (1) as below,

$  = \dfrac{{{c^4}}}{{{a^4}}} \times \left( { - \dfrac{b}{a}} \right) \times \left( {\dfrac{{{b^2} - 3ac}}{{{a^2}}}} \right) $

$   = \dfrac{{b{c^4}\left( {3ac - {b^2}} \right)}}{{{a^7}}} $

The value of $ {\alpha ^4}{\beta ^7} + {\alpha ^7}{\beta ^4} $ is $   = \dfrac{{b{c^4}\left( {3ac - {b^2}} \right)}}{{{a^7}}} $.

III. Now, we have to find the value of the given equation that is $ {\left( {\dfrac{\alpha }{\beta } - \dfrac{\beta }{\alpha }} \right)^2} $ .

For a quadratic equation $ a{x^2} + bx + c = 0 $ , the sum of its roots $ = \dfrac{{ - b}}{a} $ and the product of its roots $ = \dfrac{c}{a} $ Therefore, \[\alpha + \beta = \dfrac{{ - b}}{a}\] and $ \alpha \beta = \dfrac{c}{a} $ .

Taking the given equation

$   = {\left( {\dfrac{\alpha }{\beta } - \dfrac{\beta }{\alpha }} \right)^2} $

$   = {\left( {\dfrac{{(\alpha + \beta )(\alpha - \beta )}}{{\alpha \beta }}} \right)^2}$

$= (\alpha + \beta)^2 \dfrac{(\alpha - \beta)^2}{(\alpha \beta)^2}$

$   = \dfrac{{{b^2}}}{{{c^2}}} \times \left( {\dfrac{{{b^2} - 4ac}}{{{a^2}}}} \right) $

The value of $ {\left( {\dfrac{\alpha }{\beta } - \dfrac{\beta }{\alpha }} \right)^2} = \dfrac{{{b^2}}}{{{c^2}}} \times \left( {\dfrac{{{b^2} - 4ac}}{{{a^2}}}} \right) $.

Note:  In algebra, a quadratic equation is any equation that can be rearranged in standard form as $ ax^{2}+bx+c=0$ where x represents an unknown, and a, b, and c represent known numbers, where a ≠ 0. If a = 0, then the equation is linear, not quadratic, as there is no $ax{^2}$ term.