Welcome to Westonci.ca, where you can find answers to all your questions from a community of experienced professionals. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.
Sagot :
[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]
[tex]\rule{280pt}{2pt}[/tex]
We hope our answers were helpful. Return anytime for more information and answers to any other questions you may have. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.