Quadratic Expressions - I

Quadratic Expressions - I

Relation between roots and co-efficients
7.    Theorem:    If \( \alpha \,\,and\,\,\beta \) are the roots of ax²+bx+c=0, then prove that 
i)\( \alpha \,\, + \,\beta = \frac{{ - b}} {a} \)   ii)\( \alpha .\beta = \frac{c} {a} \) 

Proof : We know the roots of ax²+bx+c=0 are \( \frac{{ - b \pm \sqrt {b^2 - 4ac} }} {{2a}} \)
\( Let\,\,\alpha = \frac{{ - b + \sqrt {b^2 - 4ac} }} {{2a}}\,\,\,\,\,and\,\,\,\beta = \frac{{ - b - \sqrt {b^2 - 4ac} }} {{2a}} \)
i)\( \,\,\alpha + \beta = \frac{{ - b + \sqrt {b^2 - 4ac} }} {{2a}}\,\, + \frac{{ - b - \sqrt {b^2 - 4ac} }} {{2a}} \)
\( = \frac{{ - b + \sqrt {b^2 - 4ac} - b - \sqrt {b^2 - 4ac} }} {{2a}}\,\, \)
\( \begin{gathered} = \frac{{ - 2b}} {{2a}} \hfill \\ = \frac{{ - b}} {a} \hfill \\ \end{gathered} \)

ii)\( \alpha .\beta = \left( {\frac{{ - b + \sqrt {b^2 - 4ac} }} {{2a}}} \right)\left( {\frac{{ - b - \sqrt {b^2 - 4ac} }} {{2a}}} \right) \)
\( = \frac{{( - b)^2 - \left( {\sqrt {b^2 - 4ac} } \right)}} {{4a^2 }} \)
\( \begin{gathered} = \frac{{b^2 - (b^2 - 4ac)}} {{4a^2 }} \hfill \\ = \frac{{b^2 - b^2 + 4ac}} {{4a^2 }} \hfill \\ = \frac{{4ac}} {{4a^2 }} \hfill \\ = \frac{c} {a} \hfill \\ \end{gathered} \)
Hence Proved.

8. If \( \alpha \,\,and\,\,\beta \) are the roots of ax² +bx+c=0, then
i)\( \alpha \,\, + \,\beta = \frac{{ - b}} {a} \)  ii)\( \alpha \,\,.\,\beta = \frac{c} {a} \)   iii)\( \left| {\alpha \,\, - \beta } \right| = \frac{{\sqrt {b^2 - 4ac} }} {{|a|}} \)
iv)\( \alpha ^2 + \beta ^2 = \frac{{b^2 - 2ac}} {{a^2 }} \)  v)\( \alpha ^3 + \beta ^3 = \frac{{3abc - b^3 }} {{a^3 }} \)

9.The quadratic equation whose roots are
\( \alpha \,\,and\,\,\beta \,is\,(x - \alpha )\,(x - \beta ) = 0\,\,\,or\,\,x^2 - (\alpha + \beta )x + \alpha \beta = 0 \)
Note:i)\( ax^2 + bx + c = a(x - \alpha )(x - \beta ) \)
ii)\( ax^2 + bx + c = a\left( {x + \frac{b} {{2a}}} \right)^2 + \frac{{4ac - b^2 }} {{4a}} \)

Problem (i)The quadratic equation whose roots are \(1 + \sqrt 3 \,\,and\,\,1 - \sqrt 3 \)
Solution:Let \( \alpha = 1 + \sqrt 3 ,\,\,\beta = 1 - \sqrt 3 \)
we have \( \alpha + \beta = \left( {1 + \sqrt 3 } \right) + \left( {1 - \sqrt 3 } \right) = 2 \)
\( \alpha .\beta = \left( {1 + \sqrt 3 } \right).\left( {1 - \sqrt 3 } \right) = 1 - 3 = - 2 \)
The required quadratic equation is 
\( \begin{gathered} \,\,\,\,\,x^2 - \left( {\alpha + \beta } \right)x + \alpha \beta = 0 \hfill \\ \Rightarrow x^2 - (2)x + ( - 2) = 0 \hfill \\ \Rightarrow x^2 - 2x - 2 = 0 \hfill \\ \end{gathered} \)

Problem (ii)The quadratic equation whose roots are \( \frac{{2 \pm \sqrt 3 i}} {2} \) is ?
sol: Let \( \alpha = \frac{{2 + \sqrt 3 i}} {2}\,\,and\,\,\,\,\beta = \frac{{2 - \sqrt 3 i}} {2} \)
Now, \( \alpha + \beta = \frac{{2 + \sqrt {3i} }} {2}\,\, + \frac{{2 - \sqrt {3i} }} {2} \)
=\( \frac{{2 + \sqrt 3 i + 2 - \sqrt 3 i}} {2}\,\, \)
=\( \frac{4} {2} \)=2
\( \alpha .\beta = \left( {\frac{{2 + \sqrt 3 i}} {2}} \right)\,\left( {\frac{{2 - \sqrt 3 i}} {2}} \right) \)
\( \begin{gathered} = \frac{{(2)^2 - \left( {\sqrt 3 i} \right)^2 }} {4} \hfill \\ = \frac{{4 + 3}} {4}\,\,\,\,\,\,\,\,\,\,\,\,\sin ce\,\,i^2 = - 1 \hfill \\ = \frac{7} {4} \hfill \\ \end{gathered} \)
\( \therefore \) The required quadratic equation is \( x^2 - \left( {\alpha + \beta } \right)x + \alpha \beta = 0 \)
\( \begin{gathered} \Rightarrow x^2 - (2)x + \frac{7} {4} = 0 \hfill \\ \Rightarrow 4x^2 - 8x + 7 = 0 \hfill \\ \end{gathered} \)

Problem (iii) If \( \alpha ,\beta \) are the roots of ax²+bx+c=0 and , then the value of \( \frac{1} {{\left( {a\alpha + b} \right)^2 }} + \frac{1} {{\left( {a\beta + b} \right)^2 }} \) interms of a, b and c is ?
Sol : since \( \alpha \) is a root of ax²+bx+c=0 then we have \( a\alpha ^2 + b\alpha + c = 0 \)
\( \begin{gathered} \Rightarrow a\alpha ^2 + b\alpha = - c \hfill \\ \Rightarrow \alpha \left( {a\alpha + b} \right) = - c \hfill \\ \Rightarrow a\alpha + b = - \frac{c} {\alpha } \hfill \\ \end{gathered} \) 
similarly \( a\beta + b = \frac{{ - c}} {\beta } \) 
Now, \( \frac{1} {{\left( {a\alpha + b} \right)^2 }} + \frac{1} {{\left( {a\beta + b} \right)^2 }} \)
\( \begin{gathered} = \frac{1} {{\left( {\frac{{ - c}} {\alpha }} \right)^2 }} + \frac{1} {{\left( {\frac{{ - c}} {\beta }} \right)^2 }} \hfill \\ = \frac{{\alpha ^2 + \beta ^2 }} {{c^2 }} \hfill \\ \end{gathered} \)