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Math Help - Find the solution set of the equation

  1. #1
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    Red face Find the solution set of the equation

    Question : Find the solution set of the equation <br />
\begin{bmatrix}<br />
x & 3 & 7 \\<br />
2 & x & 2\\<br />
7 & 6 & x<br />
\end{bmatrix} = 0 <br />
    It is given that x= -9 is one of the roots
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  2. #2
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    Quote Originally Posted by zorro View Post
    Question : Find the solution set of the equation <br />
\begin{bmatrix}<br />
x & 3 & 7 \\<br />
2 & x & 2\\<br />
7 & 6 & x<br />
\end{bmatrix} = 0 <br />
    It is given that x= -9 is one of the roots
    Your equation does not make any sense.

    The zero matrix is <br />
\begin{bmatrix}<br />
0 & 0 & 0 \\<br />
0 & 0 & 0\\<br />
0 & 0 & 0<br />
\end{bmatrix}

    which is CLEARLY not the same as

    <br />
\begin{bmatrix}<br />
x & 3 & 7 \\<br />
2 & x & 2\\<br />
7 & 6 & x<br />
\end{bmatrix}.
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  3. #3
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    Might be this would make some sence

    Quote Originally Posted by Prove It View Post
    Your equation does not make any sense.

    The zero matrix is <br />
\begin{bmatrix}<br />
0 & 0 & 0 \\<br />
0 & 0 & 0\\<br />
0 & 0 & 0<br />
\end{bmatrix}

    which is CLEARLY not the same as

    <br />
\begin{bmatrix}<br />
x & 3 & 7 \\<br />
2 & x & 2\\<br />
7 & 6 & x<br />
\end{bmatrix}.


    Thanks fr ur reply
    Might be this might make some think clear ......The matrix is not this <br />
\begin{bmatrix}<br />
x & 3 & 7 \\<br />
2 & x & 2\\<br />
7 & 6 & x<br />
\end{bmatrix}<br />
= 0 , but this <br />
\begin{vmatrix}<br />
x & 3 & 7 \\<br />
2 & x & 2\\<br />
7 & 6 & x<br />
\end{vmatrix}<br />
= 0
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  4. #4
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    Quote Originally Posted by zorro View Post
    Thanks fr ur reply
    Might be this might make some think clear ......The matrix is not this <br />
\begin{bmatrix}<br />
x & 3 & 7 \\<br />
2 & x & 2\\<br />
7 & 6 & x<br />
\end{bmatrix}<br />
= 0 , but this <br />
\begin{vmatrix}<br />
x & 3 & 7 \\<br />
2 & x & 2\\<br />
7 & 6 & x<br />
\end{vmatrix}<br />
= 0
    The first step is to get an expression for the determinant. Then equate this expression to zero and solve for x. Please show what you've done and where you get stuck.
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  5. #5
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    The determinant will clearly be a cubic in x. Since you are already given that x= 9 is a root, you can factor it as (x-9) times some quadratic and then, if necessary, use the quadratic formula.
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  6. #6
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    Is this correct

    Quote Originally Posted by mr fantastic View Post
    The first step is to get an expression for the determinant. Then equate this expression to zero and solve for x. Please show what you've done and where you get stuck.

    After taking the deteminant this is what i have got

    x(x^2 - 12) - 3(2x - 14) + 7(12 - 7x) = 0

    x^3 - 12x - 6x + 42 + 84 - 49x = 0

    x^3 - 67x + 126 = 0

    (x + 9) (x^2 - 9x + 14)

    (x + 9) (x - 7) (x - 2)

    So what is the answer
    as x = -9 , \ 7 , \ 2

    i didnt actually understand the question therefore i dont know what to do after this ???????
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  7. #7
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    Wasn't it just asking you to solve for x?

    You've done this...
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  8. #8
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    If you want to get really, really "technical" (and mathematicians are notorious for that!), since the problem asked for "solution set", the answer is the set of those solutions: {-9, 7, 2}. (Warning: I am assuming those are the correct solutions; I didn't check them.)
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  9. #9
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    Thank u for helping me

    Quote Originally Posted by HallsofIvy View Post
    If you want to get really, really "technical" (and mathematicians are notorious for that!), since the problem asked for "solution set", the answer is the set of those solutions: {-9, 7, 2}. (Warning: I am assuming those are the correct solutions; I didn't check them.)
    Thanks HallsofIvy and Prove It for helping me

    Cheers

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