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

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

2 thoughts on “5.<br />Find the zeroes of the quadratic polynomial P(x) = x2 + x-12 and<br />verify the relationship between the zeroes and the c”

  1. 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.

  2. Answer:

    Given :

    [tex] \bf \: {x}^{2} + x – 12[/tex]

    To Find :

    Relationship the zeroes and the coefficients.

    Solution :

    We know that

    Sum of Zeroes

    [tex] \sf \: \alpha + \beta = \dfrac{ – b}{a} [/tex]

    [tex] \sf \: \alpha + \beta = \dfrac{ – ( – 1)}{1} [/tex]

    [tex] \sf \: \alpha + \beta = \dfrac{1}{1} [/tex]

    [tex] \sf \blue{ \alpha + \beta = 1}[/tex]

    Product of Zeroes

    [tex] \sf \: \alpha \beta = \dfrac{c}{a} [/tex]

    [tex] \sf \: \alpha \beta = \dfrac{ – 12}{1} [/tex]

    [tex] \alpha \beta = – 12[/tex]

    Lets factorise

    [tex] \sf \: {x}^{2} + x – 12[/tex]

    [tex] \sf \: {x}^{2} + (4x – 3x) – 12[/tex]

    [tex] \sf \: {x}^{2} + 4x – 3x + 12 = 0[/tex]

    [tex] \sf \: x(x + 4) \: or \: 3(x + 4)[/tex]

    [tex] \sf \: (x – 3) \: or \: (x + 4)[/tex]

    Either

    [tex] \bf \: {x} – 3 = 0 [/tex]

    [tex] \bf \: x \: = 0 + 3[/tex]

    [tex] \bf \: x = 3[/tex]

    Or,

    [tex] \bf \: x + 4 = 0[/tex]

    [tex] \bf \: x = 0 – 4[/tex]

    [tex] \bf \: x = – 4[/tex]

    By putting the value

    3 + (-4)

    3 – 4

    = -1

    And,

    3(-4)

    -12

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