find the equation of a quadratic with roots (1+root 5) and (1-root5) which passes through the point (2,5)?

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  • 6 months ago

    The roots are 1+√5 and 1-√5

    f(x) = a(x-(1+√5))(x-(1-√5)) = a((x-1)-√5)((x-1)+√5) =a((x-1)^2-5) = a(x^2-2x+1-5) = a(x^2-2x-4)

    5 = a(2^2-2(2)-4) = -4a

    a= -5/4

    f(x) = -5/4(x^2-2x-4)= (-5/4)x^2+(5/2)x+5

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  • 6 months ago

    y = 5/(((1+√5) - 2)((1-√5) - 2)) ((1+√5) - x)((1-√5) - x)

    y = 5/(-4) (x² - 2x - 4)

    y = -⁵∕₄ x² + ⁵∕₂ x + 5

    Graph: https://www.desmos.com/calculator/eungynfwcb

    Source(s): The equation of the polynomial curve with off x-axis point (a, b), and x intercepts x₀, x₁, etc. (not equal to a ) is y = (b/∏(xᵢ-a)) ∏(xᵢ-x).
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  • 6 months ago

    .

    Quadratic equation has a form of;

    y = ax² + bx² + c

    Using roots m and n, the quadratic equation can be derived from the following:

    y = A( x - m )( x - n )

    Distinct roots:

    x = 1 + √5 =====> ( x - 1 - √5 )

    x = 1 - √5 =====> ( x - 1 + √5 )

    Resulting equation

    y = A( x - 1 - √5 )( x - 1 + √5 )

    y = A( x² - x + x√5 - x + 1 - √5 - x√5 + √5 - 5 )

    y = A( x² - 2x - 4 )

    but passes through (2, 5)

    y = A( x² - 2x - 4 )

    5 = A( 2² - 2*2 - 4 )

    A = -5/4

    y = -5/4( x² - 2x - 4 )

    ━━━━━━━━━

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  • TomV
    Lv 7
    6 months ago

    f(x) = a((x-1)+√5)((x-1)-√5)

    = a(x-1)² - 5)

    = a(x² - 2x - 4)

    f(2) = 5

    5 = a(4 - 4 - 4)

    a = -5/4

    f(x) = -5x²/4 + 5x/2 + 5

    or

    f(x) = -5(x/2)² + 5(x/2) + 5

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  • 6 months ago

    roots (1 + √5) and (1 - √5)

    that means the equation is

    (x – (1 + √5))(x – (1 - √5)) = 0

    (x – 1 – √5)(x – 1 + √5) = 0

    x² – x + x√5 – x + 1 – √5 –x√5 + √5 – 5 = 0

    x² – 2x – 4 = 0

    ky = x² – 2x – 4

    plug in point

    k•5 = 2² – 2•2 – 4

    k•5 = 4 – 4 – 4

    k = –5/4

    so equation is

    –(5/4)y = x² – 2x – 4

    or

    –5y = 4x² – 8x – 16

    • Puzzling
      Lv 7
      6 months agoReport

      Double-check your math at the end. If you plug in x=2, you don't get y = 5.

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  • 6 months ago

    With those roots:

    y = a(x - (1 + sqrt(5)))(x - (1 - sqrt(5)))

    y = a(x - 1 - sqrt(5))(x - 1 + sqrt(5))

    y = a(x^2 - x + sqrt(5)*x - x + 1 - sqrt(5) - sqrt(5)*x + sqrt(5) - 5)

    y = a(x^2 - 2x - 4)

    Passing through that point:

    5 = a(2^2 - 2*2 - 4)

    5 = a(4 - 4 - 4)

    -4a = 5

    a = -5/4 = -1.25

    y = -1.25x^2 + 2.5x + 5

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  • mizoo
    Lv 7
    6 months ago

    f(x) = (x - (1 + √5))(x - (1 - √5))

    f(x) = (x - 1 - √5))(x - 1 + √5)

    f(x) = ((x - 1) - √5)) ((x - 1) + √5)

    f(x) = (x - 1)^2 - √5^2

    f(x) = x^2 - 2x - 4

    • ...Show all comments
    • mizoo
      Lv 7
      6 months agoReport

      I had missed the last sentence of your question. Above, I added some steps and final answer is (-5/4)(x^2 - 2x - 4). Hope this helps.

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  • ted s
    Lv 7
    6 months ago

    y = A ( ( x-1)² - 5 ) where A = - 5 / 4

    • Sally6 months agoReport

      how did u get there

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