Find the zeroes of the quadratic polynomialx2+ 7x + 12, and verify the relationship between the zeroes and coefficient​

Question

Find the zeroes of the quadratic polynomialx2+ 7x + 12, and verify the relationship between the zeroes and coefficient​

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Alice 1 month 2021-08-16T20:29:51+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-08-16T20:31:01+00:00

    Given :

    • A polynomial x²+7x+12.

    To Find :

    • Zeroes of the given polynomial and verify the relationship between zeroes and coefficient.

    Solution :

    \longmapsto\tt\bf{{x}^{2}+7x+12}

    By Splitting Middle Term :

    \longmapsto\tt{{x}^{2}+(4x+3x)+12}

    \longmapsto\tt{{x}^{2}+4x+3x+12}

    \longmapsto\tt{x(x+4)+3(x+4)}

    \longmapsto\tt{(x+3)(x+4)}

    • x = -3
    • x = -4

    So , 3 and 4 are the zeroes of polynomial x²+7x+12..

    _______________________

    Here :

    • a = 1
    • b = 7
    • c = 12

    Sum of Zeroes :

    \longmapsto\tt{\alpha+\beta=\dfrac{-b}{a}}

    \longmapsto\tt{-3+(-4)=\dfrac{-7}{1}}

    \longmapsto\tt{-3-4=7}

    \longmapsto\tt\bf{7=7}

    Product of Zeroes :

    \longmapsto\tt{\alpha\beta=\dfrac{c}{a}}

    \longmapsto\tt{-3\times{-4}=\dfrac{12}{1}}

    \longmapsto\tt\bf{12=12}

    HENCE VERIFIED

    0
    2021-08-16T20:31:12+00:00

    Given :

    A polynomial x²+7x+12.

    To Find :

    Zeroes of the given polynomial and verify the relationship between zeroes and coefficient.

    Solution :

    \longmapsto\tt\bf{{x}^{2}+7x+12}⟼x

    2

    +7x+12

    By Splitting Middle Term :

    \longmapsto\tt{{x}^{2}+(4x+3x)+12}⟼x

    2

    +(4x+3x)+12

    \longmapsto\tt{{x}^{2}+4x+3x+12}⟼x

    2

    +4x+3x+12

    \longmapsto\tt{x(x+4)+3(x+4)}⟼x(x+4)+3(x+4)

    \longmapsto\tt{(x+3)(x+4)}⟼(x+3)(x+4)

    x = -3

    x = -4

    So , -3 and -4 are the zeroes of polynomial x²+7x+12..

    _______________________

    Here :

    a = 1

    b = 7

    c = 12

    Sum of Zeroes :

    \longmapsto\tt{\alpha+\beta=\dfrac{-b}{a}}⟼α+β=

    a

    −b

    \longmapsto\tt{-3+(-4)=\dfrac{-7}{1}}⟼−3+(−4)=

    1

    −7

    \longmapsto\tt{-3-4=7}⟼−3−4=7

    \longmapsto\tt\bf{7=7}⟼7=7

    Product of Zeroes :

    \longmapsto\tt{\alpha\beta=\dfrac{c}{a}}⟼αβ=

    a

    c

    \longmapsto\tt{-3\times{-4}=\dfrac{12}{1}}⟼−3×−4=

    1

    12

    \longmapsto\tt\bf{12=12}⟼12=12

    HENCE VERIFIED

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