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Prove all the relation of quadric polynomial.

Kindly need help!!! ​


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]