Find the quadratic polynomial in each case (i) sum and product of the zeroes respectively
are _1/4 and 1/4
and (ii) for

Find the quadratic polynomial in each case (i) sum and product of the zeroes respectively
are _1/4 and 1/4
and (ii) for the zeroes a, ß are 2, – 1.​

1 thought on “Find the quadratic polynomial in each case (i) sum and product of the zeroes respectively<br />are _1/4 and 1/4<br />and (ii) for”

  1. IN THE FIRST QUESTION :

    GIVEN :

    • [tex]\alpha + \beta = \dfrac{-1}{4}[/tex]
    • [tex] \alpha \beta = \dfrac{1}{4}[/tex]

    TO FIND :

    • Quadratic Polynomial.

    FORMULA REQUIRED :

    • [tex]\underline{\boxed{\purple{\bf{x^2- \left(\alpha + \beta\right)x + \alpha\beta =0 }}}} [/tex]

    SOLUTION :

    • By substituting the values, [tex]\bf{\alpha + \beta =\dfrac{-1}{4}}[/tex] and [tex]\bf{\alpha \beta = \dfrac{1}{4}}[/tex]

    [tex] :\implies {\sf x^2 – \left(\alpha + \beta \right)x + \alpha \beta = 0}\\ \\ [/tex]

    [tex]:\implies {\sf x^2- (\dfrac{-1}{4} )x+\dfrac{1}{4} =0 }\\ \\ [/tex]

    [tex]:\implies {\sf x^2+ \dfrac{1}{4}x+\dfrac{1}{4} =0 }\\ \\ [/tex]

    [tex]:\implies {\sf \dfrac{4x^2 +x + 1}{4} =0 }\\ \\[/tex]

    • By cross multiplication :

    [tex]:\implies {\sf 4x^2+x + 1 = 0 \times 4} \\ \\ [/tex]

    [tex]:\implies{ \underline{ \boxed{ \blue{\bf{ 4 x^2 +x + 1 = 0}}}}}[/tex]

    [tex]\huge{\green{\therefore}}[/tex]Quadratic Polynomial = [tex]\bf{ 4x^2 + x + 1 }[/tex]

    IN THE SECOND QUESTION :

    GIVEN :

    • [tex]\bf{\alpha = 2}[/tex]
    • [tex]\bf{\beta = -1 }[/tex]

    TO FIND :

    • Quadratic Polynomial.

    FORMULA REQUIRED :

    • [tex]\underline{\boxed{\purple{\bf{x^2- \left(\alpha + \beta\right)x + \alpha\beta =0 }}}} [/tex]

    SOLUTION :

    Sum of Zeros :

    [tex]: \leadsto \alpha + \beta \\ \\[/tex]

    [tex]: \leadsto2 + (-1) \\ \\ [/tex]

    [tex]: \leadsto 2-1 \\ \\ [/tex]

    [tex]: \leadsto \bf{1}[/tex]

    Product of Zeros :

    [tex] : \leadsto \alpha \beta \\ \\ [/tex]

    [tex]: \leadsto 2 \times (-1)\\ \\[/tex]

    [tex]: \leadsto \bf{ -2 }[/tex]

    • By substituting the values, [tex]\bf{\alpha + \beta = 1}[/tex] and [tex]\bf{\alpha \beta = -2}[/tex]

    [tex]:\implies {\sf x^2 – \left(\alpha + \beta \right)x + \alpha \beta = 0}\\ \\ [/tex]

    [tex]:\implies {\sf x^2- \left( 1\right )x+(-2) =0 }\\ \\ [/tex]

    [tex]: \implies \underline{\boxed{ \purple{\bf{ x ^2 – x – 2 = 0 }}}}[/tex]

    [tex]\huge{\green{\therefore}}[/tex] Quadratic polynomial = [tex]\bf{ x ^2 – x -2}[/tex]

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