Mathematics

Question

For which k are the roots of k(x2+1)=x2+3x–3 real and distinct?

1 Answer

  • Answer:

    The solution for k is the interval (-3.5,1.5)

    Step-by-step explanation:

    we have

    [tex]k(x^{2}+1)=x^{2}+3x-3[/tex]

    [tex]kx^{2}+k=x^{2}+3x-3[/tex]

    [tex]x^{2}-kx^{2}+3x-3-k=0[/tex]

    [tex]}[1-k]x^{2}+3x-(3+k)=0[/tex]

    we know that

    If the discriminant is greater than zero . then the quadratic equation has two real and distinct solutions

    The discriminant is equal to

    [tex]D=b^{2}-4ac[/tex]

    In this problem we have

    a=(1-k)

    b=3

    c=-(3+k)

    substitute

    [tex]D=3^{2}-4(1-k)(-3-k)\\ \\D=9-4(-3-k+3k+k^{2})\\ \\D=9+12+4k-12k-4k^{2}\\ \\D=21-8k-4k^{2}[/tex]

    so

    [tex]21-8k-4k^{2} > 0[/tex]

    solve the quadratic equation by graphing

    The solution for k is the interval (-3.5,1.5)

    see the attached figure

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