How do I find the zeroes of this polynomial function?

p(x)=x^4-x^2+0.25 

can you show me the steps? 

13 Answers

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  • 1 month ago

    Let x^2 - y 

    Substitute 

    p(y) = y^2 - y + 0.5^2 = 0 

    Complete the Square 

    (y - 0.5)^2 - (0.5)^2 = - (0.5)^2

    (y - 0.5)^2 = -0.25 +0.25  = 0 

    ( y -0.5)^2 = 0 

    y - 0.5 = 0 

    Back substitute 

    x^2 - 0.5 = 0 

    x^2 = 0.5 

    x = +/-sqrt(0.5)

    x = +/-0.7071...   Which agrees with the graph. 

  • 1 month ago

    i was really stumped by the way the question was framed.

    as i know one equates a quadratic equation to "0",

    and tries to figure out roots.

  • 1 month ago

    It is a quadratic if you take your variable to be "x^2"

    p(x) = (x^2)^2 - (x^2) + 0.25

    use a=1, b= -1, c = 0.25

    (Remember, our variable is x^2, not x, for this to be a quadratic)

    (x^2) = [ -b ± √(b^2 - 4ac) ] / 2a

    which becomes

    (x^2) = [ 1 ± √(1 - 1) ] / 2

    (x^2) = [ 1 ± √(0) ] / 2

    x^2 = 1/2

    Therefore

    x = ± √(1/2)

    are the zeros (or the roots - in this context, it means the same thing)

  • TomV
    Lv 7
    1 month ago

    p(x) is quadratic in u if u = x². Solve the quadratic.

    u = x²

    p(u) = u² - u + 1/4 = 0

    u² - u + 1/4 = 0

    u = 1/2

    x = ±√(1/2) = ±√2/2

    Or, plot the polynomial and observe the intersections with the x-axis, x = ±0.707

    Attachment image
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  • 1 month ago
  • Amy
    Lv 7
    1 month ago

    Solving x^4 - x^2 + 0.25 = 0

    Let u = x^2

    u^2 - u + 0.25 = 0

    Apply quadratic formula.

    u = 1/2

    Convert back to x.

    x^2 = 1/2

    x = ±1/√2

  • 1 month ago

    p(x) = x⁴ - x² + 0.25

    Find the zeros, so:

    0 = x⁴ - x² + 0.25

    This is a perfect square trinomial, so:

    0 = (x² - 0.5)²

    Square root of both sides:

    0 = x² - 0.5

    x² = 0.5

    Square root of both sides again:

    x = ± √(0.5)

    Let's simplify that a little.  If we turn that into a fraction we can then rationalize the denominator:

    x = ± √(1/2)

    x = ± 1/√2

    x = ± √(2)/2

    or:

    x = ±(1/2)√2

  • 1 month ago

    I hope this helps. Use substitution and then quadratic equation

    Attachment image
  • 1 month ago

    p(x) = x^4 - x^2 + 0.25 

    p(x) = (x - 0.707107)^2 (x + 0.707107)^2

    Roots:

    x ≈ -0.707107

    x ≈ 0.707107

  • 1 month ago

    x⁴ – x² + 1/4 = 0

    substitute z = x²

    z² – z  + 1/4 = 0

    quadratic equation:to solve ax² + bx + c = 0x = [–b ± √(b²–4ac)] / 2az = [1 ± √(1–1] / 2

    z = 1/2

    z = x²

    x = ±√(1/2)

    check

    x⁴ – x² + 1/4 = 0

    x = √(1/2)

    (√(1/2))⁴ – (√(1/2))² + 1/4 = 0

    1/4 – 1/2 + 1/4 = 0

    ok

    x = –√(1/2)

    (–√(1/2))⁴ – (–√(1/2))² + 1/4 = 0

    1/4 – 1/2 + 1/4 = 0

    ok

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