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Given that f(x)= ax2+bx+c has two …

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Given that f(x)= ax2+bx+c has two zeroes -2 and 1/2 . It leaves a remaunder 12 when divided by x-2 , find thevalue of a,b and c

    • Posted by Subhjot Kaur 5 years, 11 months ago

    CBSE > Class 10 > Mathematics
    • 7 answers

    Aseem Mahajan 5 years, 11 months ago

    From (i) , (ii) ,(iii) a=2 , b=3 , c=-2

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    Aseem Mahajan 5 years, 11 months ago

    f(2) = a(2)²+b(2)+c = 12 => 4a+2b+c=12 ---------(iii)

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    Aseem Mahajan 5 years, 11 months ago

    When divided by x-2 , remainder is 12 ,, so f(2) = 12

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    Aseem Mahajan 5 years, 11 months ago

    f(1/2) = a(1/2)²+b(1/2)+c => a/4+b/2+c =0 => a+2b+4c=0 -------(ii)

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    Aseem Mahajan 5 years, 11 months ago

    4a* not 4a² ..

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    Aseem Mahajan 5 years, 11 months ago

    f(-2) = a(-2)²+b(-2)+c => 4a²-2b+c=0 -----(i)

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    Aseem Mahajan 5 years, 11 months ago

    If -2 and 1/2 are zeroes of f(x), then f(-2) = f(1/2) = 0

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