Anonymous
Anonymous asked in Science & MathematicsMathematics · 9 years ago

Show that the following equation has real roots for all values of k ?

Show that the equation (k-1)x^2 + 2x - (k-3) = 0

has real roots for all values of k.

Please go through step.

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  • 9 years ago
    Favorite Answer

    (k-1)x^2+2x-(k-3)=0

    suppose k-1=a and k-3=b

    so,a-b=k-1-k+3=2

    so the equation becomes ax^2+(a-b)x-b=0

    => ax^2+ax-bx-b=0

    =>ax(x+1)-b(x+1)=0

    =>(ax-b)(x-1)=0

    x=-1 or x=b/a=(k-3)/(k-1)

  • Matt
    Lv 4
    9 years ago

    use the quadratic equation: (-b±√(b^2-4ac))/2a where a=(k-1), b=2, c=-(k-3)

    (2±√(4-4(k-1)-(k-3)))2(k-1)

    Factor out a 4 and multiply out under the square root sign:

    (2±√(4(1+k^2-4k+3)))/2(k-1)

    Now, square root of 4 is 2, and putting together like terms you have:

    (2±2√(k^2-4k+4))/2(k+1)

    Now looking at just what is under the square root sign: k^2-4k+4, which will factor to (k-2)^2.

    square root of a square, so you have (2±2(k-2))/2(k-1)

    Factor out a 2: (1±(k-2))/(k-1)

    So, since we were able to factor out of the square root sign without getting a negative number, there can be only real roots.

    BUT at the end you have a (k-1) in the denominator. so if k=1, we do not have an answer. BUT now look at the original equation. When k=1, the x^2 term becomes 0 and we are left with 2x-(1-3)=0, or 2x+2=0 or x=-1. So even when k=1, we still have a real root.

    Therefore for all values of k, we will have real roots.

  • 9 years ago

    If a quadratic equation has real roots, the the discriminant must be = 0 or > 0 ...;

    The discriminant is b^2 - 4ac

    Then in your case b^2 -4ac = 2^2 -4(k-1)[-(k-3)] ... and for real roots

    4 +4(k-1)(k-3)>= 0 ==> 4 + 4(k^2 +3k + 3)>=0 here k^2 +3k +3 has no real root, that is the graph does not cut the x axis... it is positive for all value of K ... the the discriminant is always positive for all value of K. OK!

  • 9 years ago

    (k-1)x^2 + 2x - (k-3) = 0

    Discriminator = 2^2 + 4(k-1)(k-3) =

    = 4+4k^2-16k+12 = 4(k*2 -4k +4) =

    = 4(k -2)^2 ....................................... [1]

    Analysing [1], the equation has real roots for all values of k.

    <============ A . N . S . W . E . R ================>

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  • ?
    Lv 7
    9 years ago

    (k-1)x^2 + 2x - (k-3) = 0

    quadratic equation says that for general equation ax^2+bx+c=0

    solution is

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

    b^2 - 4ac is the important part to determine the nature of the roots (known as the discriminant)

    In our case we can see that the discriminant, D = 4 + 4(k-1)(k-3) = 4(k-2)^2

    Therefore, we can now see that for all values of k, D will be a positive real number. ie all roots are real.

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