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Thread: value of m

  1. #1
    rcs
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    value of m

    solve for x^2 - mx - 5 whose sum of roots is 6. find m.

    can any body help me on this ?

    thanks
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    Re: value of m

    Quote Originally Posted by rcs View Post
    solve for x^2 - mx - 5 whose sum of roots is 6. find m.
    can any body help me on this ?
    $\displaystyle (x-q)(x-p)=x^2-(p+q)x+p\cdot q=0$.

    Do you see the sum of the roots?
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  3. #3
    rcs
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    Re: value of m

    Quote Originally Posted by Plato View Post
    $\displaystyle (x-q)(x-p)=x^2-(p+q)x+p\cdot q=0$.

    Do you see the sum of the roots?
    so therefore p+q = 6 ... how is it possible to get p x q = 5? ...
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    Re: value of m

    Quote Originally Posted by rcs View Post
    how is it possible to get p x q = 5? ...

    It is not possible because $\displaystyle p\cdot q=-5~.$
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  5. #5
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    Re: value of m

    Hello, rcs!

    $\displaystyle \text{The sum of roots of }\,x^2 - mx - 5\:=\:0\,\text{ is 6. Find }m.$

    Suppose the two roots are $\displaystyle p$ and $\displaystyle q$.

    Vieta's formulas tell us: .$\displaystyle p+q \:=\:m,\quad pq \:=\:-5$

    . . Therefore: .$\displaystyle m \,=\,6$


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    The equation becomes: .$\displaystyle x^2 - 6x - 5 \:=\:0$

    Quadratic Formula: .$\displaystyle x \;=\;\frac{6 \pm\sqrt{36+20}}{2} \;=\;3 \pm\sqrt{14} $

    Hence: .$\displaystyle \begin{Bmatrix}p &=& 3 + \sqrt{14} \\ q &=& 3 - \sqrt{14}\end{Bmatrix}$


    Check:

    $\displaystyle p+q \;=\;(3+\sqrt{14}) + (3 - \sqrt{14}) \;=\;6\;\;\checkmark $

    $\displaystyle pq \;=\;(3 + \sqrt{14})(3-\sqrt{14}) \;=\;9 - 14 \;=\;-5\;\;\checkmark $
    Thanks from rcs
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  6. #6
    rcs
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    Re: value of m

    Wow i got it. Thanks.... My god this is magic ... Thanks Sir Soroban
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