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Prove that x2+4x+5 has no real …

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Prove that x2+4x+5 has no real zeroes

    • Posted by Devanshi Verma 6 years, 6 months ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    Sia ? 6 years, 6 months ago

    f(x) = x4 + 4x2 + 5

    = (x2)2 + 4x2 + 5
    Let x2 =n,
    Then, f(x) = n2 + 4n + 5,

    Here a=1,b=4,c=5

    The discriminant(D) = {tex}\text{b}^2-4\mathrm{ac}=\;(4)^2-4\times1\times5=16-20=-4{/tex}

    Since the discriminant is negative so this polynomial has no zeros

    Hence, f(x) = x4 + 4x2 + 5 has no zero.

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    CBSE > Class 10 > Mathematics
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