LMN asked in Science & MathematicsMathematics · 1 decade ago

Prove that, for all real values of k, the roots of the quadratic equation x^2-(2+k)x-3=0 are real.?

show further that the roots are of opposite signs. Does anyone know how to do this???? I'd appreciate any help!!

2 Answers

  • ?
    Lv 4
    1 decade ago
    Best Answer

    You approach a question like this by using the quadratic equation (for quadratic forms like ax^2+bx+c) and in particular the sqrt in the quadratic equation, which is called the discriminant:

    discriminant: b^2-4ac

    Now when the discriminant is positive there are 2 real solutions to the equation, when the discriminant is equal to zero there is 1 real solution, and when the discriminant is negative there are no solutions to the equation (this is because in the quadratic formula the discriminant is under a sqrt sign and you cannot take the sqrt of a negative number - thus no solution is possible).

    2 solutions: discriminant>0

    1 solution: discriminant=0

    0 solutions: discriminant<0

    In our question our a, b, & c are as follows:





    [-(2+k)]^2 - 4(1)(-3) = (k+2)^2+12 > 0 (always positive as a square plus a positive number is positive)

    after expanding:

    =k^2+4k+16 which is always positive (this can be shown by graphing or by again computing the discriminant and showing this new quadratic has no solutions -> meaning it never crosses the x-axis and is always positive).

  • mchale
    Lv 4
    3 years ago

    If the quadratic equation ax^2+bx+c = 0 has no genuine roots, then b^2 - 4ac < 0 With 7x^2 + kx + 8 = 0, b^2 - 4ac = ok^2 - 4(7)(8) = ok^2 - 224 for this reason for no genuine roots, ok^2 - 224 < 0 i.e. -4?14 < ok < 4?14 or ok lies interior the era (-4?14, 4?14)

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