5.
Find the zeroes of the quadratic polynomial P(x) = x2 + x-12 and
verify the relationship between the zeroes and the c

Question

5.
Find the zeroes of the quadratic polynomial P(x) = x2 + x-12 and
verify the relationship between the zeroes and the coefficients.​

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Ximena 3 months 2021-07-06T15:31:15+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-07-06T15:32:53+00:00

    EXPLANATION.

    Quadratic equation.

    ⇒ x² + x – 12.

    As we know that,

    Sum of the zeroes of the quadratic equation.

    ⇒ α + β = -b/a.

    ⇒ α + β = -(1)/1 = -1.

    Products of the zeroes of the quadratic equation.

    ⇒ αβ = c/a.

    ⇒ αβ = (-12)/1 = -12.

    As we know that,

    Factorizes the equation into middle term splits, we get.

    ⇒ x² + x – 12 = 0.

    ⇒ x² + 4x – 3x – 12 = 0.

    ⇒ x(x + 4) – 3(x + 4) = 0.

    ⇒ (x – 3)(x + 4) = 0.

    ⇒ x = 3 and x = -4.

    Sum of the values of x, we get.

    ⇒ 3 + (-4) = -1.

    Products of the values of x, we get.

    ⇒ (3)(-4) = -12.

                                                                                                                                 

    MORE INFORMATION.

    Conditions for common roots.

    Let quadratic equation are a₁x² + b₁x + c₁ = 0 and a₂x² + b₂x + c₂ = 0.

    (1) = If only one roots is common.

    ⇒ x = b₁c₂ – b₂c₁/a₁b₂ – a₂b₁.

    ⇒ y = c₁a₂ – c₂a₁/a₁b₂ – a₂b₁.

    (2) = If both roots are common.

    ⇒ a₁/a₂ = b₁/b₂ = c₁/c₂.

    0
    2021-07-06T15:33:09+00:00

    Answer:

    Given :

     \bf \:  {x}^{2}  + x - 12

    To Find :

    Relationship the zeroes and the coefficients.

    Solution :

    We know that

    Sum of Zeroes

     \sf \:  \alpha  +  \beta  =  \dfrac{ - b}{a}

     \sf \:  \alpha  +  \beta  =  \dfrac{ - ( - 1)}{1}

     \sf \:  \alpha  +  \beta  =  \dfrac{1}{1}

     \sf \blue{ \alpha  +  \beta = 1}

    Product of Zeroes

     \sf \:  \alpha  \beta  =  \dfrac{c}{a}

     \sf \:  \alpha  \beta  =  \dfrac{ - 12}{1}

     \alpha  \beta  =  - 12

    Lets factorise

     \sf \:  {x}^{2}  + x - 12

     \sf \:  {x}^{2}  + (4x - 3x) - 12

     \sf \:  {x}^{2}  + 4x - 3x + 12 = 0

     \sf \: x(x + 4) \: or \: 3(x + 4)

     \sf \: (x - 3) \: or \: (x + 4)

    Either

     \bf \:  {x} - 3 = 0

     \bf \: x \:  = 0 + 3

     \bf \: x = 3

    Or,

     \bf \: x  + 4 = 0

     \bf \: x = 0 - 4

     \bf \: x =  - 4

    By putting the value

    3 + (-4)

    3 – 4

    = -1

    And,

    3(-4)

    -12

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