Question

Prove all the relation of quadric polynomial.

Kindly need help!!! ​

Answer 1

`{\large{\textsf{\textbf{\underline{\underline{Solution :}}}}}}`

Let
`\alpha` and
`\beta` be the two zeroes of P(x) =
`\sf {a}^(2) + bx + c`

• A polynomial is always equal to it's factors, also the constant "k" is not equal to zero

`\tt \sf {a}^(2) + bx + c = k(x - \alpha )(x - \beta )`

• Using distributive property

`\tt \sf {a}^(2) + bx + c = k \bigg( {x}^(2) - \beta x - \alpha x + \alpha \beta \bigg)`

`\tt \sf {a}^(2) + bx + c = k {x}^(2) -k (\beta x )- k(\alpha x )+k( \alpha \beta )`

• Taking common

`\tt \sf {a}^(2) + bx + c = k {x}^(2) -kx (\beta + \alpha)+k( \alpha \beta ) - - (1)`

• Equating coefficients of like terms

`\sf \: a = k - - (2)`

`\sf b = - k( \alpha + \beta ) - - (3)`

`\sf c = k \alpha \beta - - (4)`

★ From 3

`\sf b = - k( \alpha + \beta ) - - (3)`

★ From 2 we have a = k so we have -

`\sf b = - a( \alpha + \beta )`

`\sf - (b)/(a) = ( \alpha + \beta )`

`\sf \therefore \boxed {{ \red{( \alpha + \beta ) = - (b)/(a) } }}`

• Sum of zeros =
`- (b)/(a)`

★ From 4

`\sf c = k \alpha \beta - - (4)`

★ From 2 we have a = k so we have -

`\sf c = a \: \alpha \: \beta`

`\sf (c)/(a) = \alpha \beta`

`\sf \therefore \boxed {{ \red{( \alpha \beta ) = (c)/(a) } }}`

• Product of zeros =
`(c)/(a)`

Also,

★ From 1

`\tt \sf {a}^(2) + bx + c = k {x}^(2) -kx (\beta + \alpha)+k( \alpha \beta ) - - (1)`

`\tt \sf {a}^(2) + bx + c = k \bigg( {x}^(2) -x ( \alpha + \beta ) + ( \alpha \beta \bigg)`

• Now just put S instead of sum of zeroes and P instead of product of zeroes

`\tt \sf {a}^(2) + bx + c = k \bigg( {x}^(2) -x ( S ) + ( P \bigg)`

`\tt \sf{a}^(2) + bx + c = \boxed{ \red{ \sf \tt k \bigg( {x}^(2) - Sx + ( P \bigg ) }}`

`\rule{280pt}{2pt}`