Find the set of values of k?

Find the set of values of k for which the line y = 2x + k cuts the curve y = x2 + kx + 5 at two

distinct points.

PLease include simple explanations on each steps.

3 Answers

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  • 4 years ago
    Best Answer

    y = 2x + k ← this is a line

    y = x² + kx + 5 ← this is a curve

    y = y → 2 points → 2 solutions for x

    x² + kx + 5 = 2x + k

    x² + kx + 5 - 2x - k = 0

    x² + x.(k - 2) + (5 - k) = 0

    Polynomial like: ax² + bx + c, where:

    a = 1

    b = (k - 2)

    c = (5 - k)

    Δ = b² - 4ac (discriminant)

    Δ = (k - 2)² - 4.(5 - k)

    Δ = k² - 4k + 4 - 20 + 4k

    Δ = k² - 16

    Δ = (k + 4).(k - 4) → 2 solutions for x → Δ > 0 → the roots are (- 4), (4) → then you make a table

    k_____-∞____- 4____4____+∞

    (k + 4)_____-__0__+____+

    (k - 4)______-____-__0__+

    Δ sign_____+__II_-__0__+

    …and you can see when the sign is > 0

    → k Є ] -∞ ; - 4 [ U ] 4 ; +∞ [

    x₁ = (- b - √Δ)/2a → y₁ = 2x₁ + k

    x₂ = (- b + √Δ)/2a → y₂ = 2x₂ + k

  • Raj K
    Lv 7
    4 years ago

    y = 2x + k cuts the curve y = x2 + kx + 5

    → 2x + k=x2 + kx + 5

    or x²+(k−2)x+5−k=0

    Hence a=1, b=k−2 and c=5−k

    Discriminant D=b²−4ac=(k−2)²−4×1×(5−k) > 0

    →k²−4k+4−20+4k> 0

    i.e. →k²−16> 0

    k²>16 or either k > 4 or k < −4

    Hence set of values is (−∞,−4)U(4,∞ )

  • cidyah
    Lv 7
    4 years ago

    y=2x+k

    y= x^2+kx+5

    x^2+kx+5 = 2x+k

    x^2+kx-2x+5-k=0

    x^2+x(k-2)+(5-k)= 0

    ax^2+bx+c = 0

    a=1

    b=k-2

    c=5-k

    discriminant = b^2-4ac > 0 for real roots

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

    = k^2-4k+4-20+4k

    = k^2-16 > 0

    solve k^2-16 =0

    (k+4)(k-4)=0

    k=-4,4

    Consider the intervals:

    (-∞ ,-4),(-4,4),(4, ∞)

    choose any one point from each interval and examine the sign of k^2-16

    (-∞ ,-4): choose k=-5

    k^2-16 = 9 > 0 (a solution)

    (-4,4) : choose k = 0

    k^2-16 = -16 < 0 (solution does not hold)

    (4, ∞): choose k= 5

    k^2-16 = 9 > 0 (solution)

    Therefore: k < -4 and k > 4 are solutions

    (-∞ ,-4) U (4, ∞)

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