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Math Help - value of k and number of root

  1. #1
    rcs
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    value of k and number of root

    how is k solved when 4x^2 + kx + 6 = 0...

    i really get so hard to solve when there is already k in the equation.
    kindly help me.

    Thanks a lot so much.
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    Re: value of k and number of root

    As is you cannot solve for k.
    To find k you'd need to know how many real roots the equation has and then use the discriminant as appropriate.

    Alternatively you can find x in terms of k using your favourite method
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    Re: value of k and number of root

    Quote Originally Posted by rcs View Post
    how is k solved when 4x^2 + kx + 6 = 0...

    i really get so hard to solve when there is already k in the equation.
    kindly help me.

    Thanks a lot so much.
    what the discriminant tells you ...

    b^2 - 4ac < 0 no real roots

    b^2 - 4ac = 0 one real root of multiplicity two

    b^2 - 4ac > 0 two real roots

    so, if you're looking for the value of k that provides at least one real root ...

    k^2 - 4(4)(6) \ge 0

    solve for k
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  4. #4
    rcs
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    Re: value of k and number of root

    thanks... it this correct sir: k > = +- 4 (sqrt 6), is this the root now sir?
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    Re: value of k and number of root

    Almost: |k| \geq 4\sqrt{6} is true.

    It follows that k \geq 4\sqrt6 which is one of your solutions yet we need to look at the other root: -k \geq 4\sqrt{6} \longrightarrow \ k \leq -4\sqrt6
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    Re: value of k and number of root

    this one is not so bad.

    if we use the idea of comparing coefficients as has been shown you can think like this.

    (A_1x+B_1)(A_2x+B_2)=A_1A_2x^2+A_1B_2x+A_2B_1x+B_1  B_2

    now you have 4x^2+kx+6 all positive terms. the coefficients are 4 and 6. which factor to A_1A_2=4=2\cdot 2\cdot 1 and B_1B_2=6=3\cdot 2\cdot 1. so you have 2 factors of each coeffient just like the general form. now just substitute all the inormation into the general factors.

    (A_1x+B_1)(A_2x+B_2) \rightarrow (2x+2)(2x+3) \rightarrow 4x^2+10x+6

    k=10

    now that you know k solve your linear equations to find the roots.
    Last edited by skoker; July 10th 2011 at 08:20 PM.
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  7. #7
    rcs
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    Re: value of k and number of root

    skoker... im kind of confuse... how can you relate k = 10 from the answer given by e^(i*pi) above? how is it possible.. i think they are not the same

    thanks
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    Re: value of k and number of root

    I think I get it right no?

    k=10 roots= -1 , -3/2.

    I'm not sure what kind of business they have gotten into up there...
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    Re: value of k and number of root

    here is a extra think...

    to check k I do this.
    b^2-4ac>0 \rightarrow 10^2-4\cdot 4\cdot \6>0 \rightarrow 4>0 2 real roots.

    now if we do like before we get this.
    k^2-4ac>0 \rightarrow k^2-4\cdot 4 \cdot \6>0 \rightarrow k^2-96>0 \rightarrow k^2>96 \rightarrow k>\pm 4 \sqrt{6}

    now we check with k and decimals. \pm 4 \sqrt{6}=\pm 9.7980 \;,\; k=10
    10>9.7980 \;,\; 10>-9.7980

    the inequalities is only telling you what value k is greater then or less then not k itself.
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  10. #10
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    Re: value of k and number of root

    To clarify mine and Skeeter's answers were stating the values of k for which there would be at least one real root.

    Skoker has given the actual value of k. For what it's worth I'm confused as to how (A_1x + B_1)(A_2x + B_2) becomes (2x+2)(2x+3)
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    Re: value of k and number of root

    Quote Originally Posted by e^(i*pi) View Post
    I'm confused as to how (A_1x + B_1)(A_2x + B_2) becomes (2x+2)(2x+3)
    I think you guys over think on this one.

    for quadratic equations this is true. all partial products of factors.
    (A_1x+B_1)(A_2x+B_2)=A_1A_2x^2+A_1B_2x+A_2B_1x+B_1  B_2

    and we have the rule ax^2+bx+c \;,\; a_1c_1+a_2c_2=b in this case a and c only have 2 factors each so they fall nicely into the LHS but you could use the same method for any ac coefficients that factor.

    A_1=2 , A_2=2 , B_1=2 , B_2=3
    Last edited by skoker; July 11th 2011 at 07:25 AM.
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