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[tex]{\large{\textsf{\textbf{\underline{\underline{Solution :}}}}}}[/tex]
Let [tex]\alpha[/tex] and [tex]\beta[/tex] be the two zeroes of P(x) = [tex]\sf {a}^{2} + bx + c[/tex]
• A polynomial is always equal to it's factors, also the constant "k" is not equal to zero
[tex] \tt \sf {a}^{2} + bx + c = k(x - \alpha )(x - \beta )[/tex]
• Using distributive property
[tex]\tt \sf {a}^{2} + bx + c = k \bigg( {x}^{2} - \beta x - \alpha x + \alpha \beta \bigg)[/tex]
[tex]\tt \sf {a}^{2} + bx + c = k {x}^{2} -k (\beta x )- k(\alpha x )+k( \alpha \beta )[/tex]
• Taking common
[tex]\tt \sf {a}^{2} + bx + c = k {x}^{2} -kx (\beta + \alpha)+k( \alpha \beta ) - - (1)[/tex]
• Equating coefficients of like terms
[tex] \sf \: a = k - - (2)[/tex]
[tex] \sf b = - k( \alpha + \beta ) - - (3)[/tex]
[tex] \sf c = k \alpha \beta - - (4)[/tex]
★ From 3
[tex] \sf b = - k( \alpha + \beta ) - - (3)[/tex]
★ From 2 we have a = k so we have -
[tex] \sf b = - a( \alpha + \beta ) [/tex]
[tex]\sf - \dfrac{b}{a} = ( \alpha + \beta ) [/tex]
[tex]\sf \therefore \boxed {{ \red{( \alpha + \beta ) = - \dfrac{b}{a} } }}[/tex]
• Sum of zeros = [tex]- \dfrac{b}{a}[/tex]
★ From 4
[tex] \sf c = k \alpha \beta - - (4)[/tex]
★ From 2 we have a = k so we have -
[tex] \sf c = a \: \alpha \: \beta[/tex]
[tex]\sf \dfrac{c}{a} = \alpha \beta[/tex]
[tex]\sf \therefore \boxed {{ \red{( \alpha \beta ) = \dfrac{c}{a} } }}[/tex]
• Product of zeros = [tex] \dfrac{c}{a}[/tex]
Also,
★ From 1
[tex]\tt \sf {a}^{2} + bx + c = k {x}^{2} -kx (\beta + \alpha)+k( \alpha \beta ) - - (1)[/tex]
[tex]\tt \sf {a}^{2} + bx + c = k \bigg( {x}^{2} -x ( \alpha + \beta ) + ( \alpha \beta \bigg) [/tex]
• Now just put S instead of sum of zeroes and P instead of product of zeroes
[tex] \tt \sf {a}^{2} + bx + c = k \bigg( {x}^{2} -x ( S ) + ( P \bigg) [/tex]
[tex]\tt \sf{a}^{2} + bx + c = \boxed{ \red{ \sf \tt k \bigg( {x}^{2} - Sx + ( P \bigg ) }}[/tex]
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