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Help with Discriminant(C1)

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    I have this C1 discriminant question that I can't do, anyone know how to do it?
    Prove that the roots of the equation kx^2+(2k+4)x+8 are real for all values of K.
    x^2=x squared
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    (Original post by Josh_J_Mead_)
    I have this C1 discriminant question that I can't do, anyone know how to do it?
    Prove that the roots of the equation kx^2+(2k+4)x+8 are real for all values of K.
    x^2=x squared
    So show that b^2-4ac\geq0 for all k \in \mathbb{R}

    a=k

    b=2k+4

    c=8
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    Boring part

    For the equation ax^2 + bx + c = 0, the discriminant is b^2-4ac, and if this number is positive then there are two real roots. So you can figure out what a, b and c are in this particular equation (don't let the ks put you off, the process is the same), then show that this must be positive for real k.

    Interesting part

    Why is this true?

    Because when you complete the square, you get

    x^2 + bx/a + c/a = 0

    (x + b/2a)^2 - b^2/4a^2 + c/a = 0

    (x + b/2a)^2 = b^2/4a^2 - c/a

    = b^2/4a^2 - 4ac/4a^2
    = (b^2 - 4ac)/(4a^2)

    So

    (x + b/2a)^2 = (b^2 - 4ac)/(4a^2)
    4a^2*(x + b/2a)^2 = (b^2 - 4ac)
    [2a*(x + b/2a)]^2 = b^2 - 4ac

    2a*(x + b/2a) = +/- sqrt(b^2 - 4ac)

    That last line tells you why it is true. If b^2-4ac is positive then its square root, the RHS of this equation, is real and we have real solutions for x. If it is negative then the RHS is an imaginary number, and we have no real solutions. So b^2 - 4ac is the key.
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    Thank you
    (Original post by mik1a)
    Boring part

    For the equation ax^2 + bx + c = 0, the discriminant is b^2-4ac, and if this number is positive then there are two real roots. So you can figure out what a, b and c are in this particular equation (don't let the ks put you off, the process is the same), then show that this must be positive for real k.

    Interesting part

    Why is this true?

    Because when you complete the square, you get

    x^2 + bx/a + c/a = 0

    (x + b/2a)^2 - b^2/4a^2 + c/a = 0

    (x + b/2a)^2 = b^2/4a^2 - c/a

    = b^2/4a^2 - 4ac/4a^2
    = (b^2 - 4ac)/(4a^2)

    So

    (x + b/2a)^2 = (b^2 - 4ac)/(4a^2)
    4a^2*(x + b/2a)^2 = (b^2 - 4ac)
    [2a*(x + b/2a)]^2 = b^2 - 4ac

    2a*(x + b/2a) = +/- sqrt(b^2 - 4ac)

    That last line tells you why it is true. If b^2-4ac is positive then its square root, the RHS of this equation, is real and we have real solutions for x. If it is negative then the RHS is an imaginary number, and we have no real solutions. So b^2 - 4ac is the key.

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