Solve second degree polynomial by “completing the square”












0












$begingroup$


I can find the first $x$, but not the other. I know that the other $x$ is equal to $-4$.



$$begin{align*}
1 &= (x^2+2x+1)-8\
9&=(x^2+2x+1)\
9&= (x_1+1)^2\
3&= x_1+1\
2&= x_1
end{align*}$$



How do I deduce what the other $x$ is by using the method "completing the square"?






algebra-precalculus polynomials







share|cite|improve this question











$endgroup$












  • $begingroup$
    The inverse of squaring is $pm$ square root. You forgot to include the negative version when you undid the square.
    $endgroup$
    – Adrian Keister
    Jan 17 at 21:12










  • $begingroup$
    $9=left(x_1+1right)^2implies sqrt{9}=sqrt{left(x_1+1right)^2}impliespm3=x_1+1 $
    $endgroup$
    – coreyman317
    Jan 17 at 21:13






  • 2




    $begingroup$
    You need to remember that $sqrt{a^2}=|a|$. So the fourth line shuld read $3=|x+1|$, which means that either you have $3=x+1$, or you have $-3=x+1$.
    $endgroup$
    – Arturo Magidin
    Jan 17 at 21:14


















0












$begingroup$


I can find the first $x$, but not the other. I know that the other $x$ is equal to $-4$.



$$begin{align*}
1 &= (x^2+2x+1)-8\
9&=(x^2+2x+1)\
9&= (x_1+1)^2\
3&= x_1+1\
2&= x_1
end{align*}$$



How do I deduce what the other $x$ is by using the method "completing the square"?






algebra-precalculus polynomials







share|cite|improve this question











$endgroup$












  • $begingroup$
    The inverse of squaring is $pm$ square root. You forgot to include the negative version when you undid the square.
    $endgroup$
    – Adrian Keister
    Jan 17 at 21:12










  • $begingroup$
    $9=left(x_1+1right)^2implies sqrt{9}=sqrt{left(x_1+1right)^2}impliespm3=x_1+1 $
    $endgroup$
    – coreyman317
    Jan 17 at 21:13






  • 2




    $begingroup$
    You need to remember that $sqrt{a^2}=|a|$. So the fourth line shuld read $3=|x+1|$, which means that either you have $3=x+1$, or you have $-3=x+1$.
    $endgroup$
    – Arturo Magidin
    Jan 17 at 21:14
















0












0








0





$begingroup$


I can find the first $x$, but not the other. I know that the other $x$ is equal to $-4$.



$$begin{align*}
1 &= (x^2+2x+1)-8\
9&=(x^2+2x+1)\
9&= (x_1+1)^2\
3&= x_1+1\
2&= x_1
end{align*}$$



How do I deduce what the other $x$ is by using the method "completing the square"?






algebra-precalculus polynomials







share|cite|improve this question











$endgroup$




I can find the first $x$, but not the other. I know that the other $x$ is equal to $-4$.



$$begin{align*}
1 &= (x^2+2x+1)-8\
9&=(x^2+2x+1)\
9&= (x_1+1)^2\
3&= x_1+1\
2&= x_1
end{align*}$$



How do I deduce what the other $x$ is by using the method "completing the square"?






algebra-precalculus polynomials




algebra-precalculus polynomials






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 17 at 21:13









Arturo Magidin

264k34587915




264k34587915










asked Jan 17 at 21:10









Ryan CameronRyan Cameron

697




697












  • $begingroup$
    The inverse of squaring is $pm$ square root. You forgot to include the negative version when you undid the square.
    $endgroup$
    – Adrian Keister
    Jan 17 at 21:12










  • $begingroup$
    $9=left(x_1+1right)^2implies sqrt{9}=sqrt{left(x_1+1right)^2}impliespm3=x_1+1 $
    $endgroup$
    – coreyman317
    Jan 17 at 21:13






  • 2




    $begingroup$
    You need to remember that $sqrt{a^2}=|a|$. So the fourth line shuld read $3=|x+1|$, which means that either you have $3=x+1$, or you have $-3=x+1$.
    $endgroup$
    – Arturo Magidin
    Jan 17 at 21:14




















  • $begingroup$
    The inverse of squaring is $pm$ square root. You forgot to include the negative version when you undid the square.
    $endgroup$
    – Adrian Keister
    Jan 17 at 21:12










  • $begingroup$
    $9=left(x_1+1right)^2implies sqrt{9}=sqrt{left(x_1+1right)^2}impliespm3=x_1+1 $
    $endgroup$
    – coreyman317
    Jan 17 at 21:13






  • 2




    $begingroup$
    You need to remember that $sqrt{a^2}=|a|$. So the fourth line shuld read $3=|x+1|$, which means that either you have $3=x+1$, or you have $-3=x+1$.
    $endgroup$
    – Arturo Magidin
    Jan 17 at 21:14


















$begingroup$
The inverse of squaring is $pm$ square root. You forgot to include the negative version when you undid the square.
$endgroup$
– Adrian Keister
Jan 17 at 21:12




$begingroup$
The inverse of squaring is $pm$ square root. You forgot to include the negative version when you undid the square.
$endgroup$
– Adrian Keister
Jan 17 at 21:12












$begingroup$
$9=left(x_1+1right)^2implies sqrt{9}=sqrt{left(x_1+1right)^2}impliespm3=x_1+1 $
$endgroup$
– coreyman317
Jan 17 at 21:13




$begingroup$
$9=left(x_1+1right)^2implies sqrt{9}=sqrt{left(x_1+1right)^2}impliespm3=x_1+1 $
$endgroup$
– coreyman317
Jan 17 at 21:13




2




2




$begingroup$
You need to remember that $sqrt{a^2}=|a|$. So the fourth line shuld read $3=|x+1|$, which means that either you have $3=x+1$, or you have $-3=x+1$.
$endgroup$
– Arturo Magidin
Jan 17 at 21:14






$begingroup$
You need to remember that $sqrt{a^2}=|a|$. So the fourth line shuld read $3=|x+1|$, which means that either you have $3=x+1$, or you have $-3=x+1$.
$endgroup$
– Arturo Magidin
Jan 17 at 21:14












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Caution:



$$(x+1)^2=9implies x+1=color{red}{pm 3}impliesbegin{cases}x_1+1=-3implies x_1=-4\{}\x_2+1=3implies x_2=2end{cases}$$






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    $begingroup$

    Caution:



    $$(x+1)^2=9implies x+1=color{red}{pm 3}impliesbegin{cases}x_1+1=-3implies x_1=-4\{}\x_2+1=3implies x_2=2end{cases}$$






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      $begingroup$

      Caution:



      $$(x+1)^2=9implies x+1=color{red}{pm 3}impliesbegin{cases}x_1+1=-3implies x_1=-4\{}\x_2+1=3implies x_2=2end{cases}$$






      share|cite|improve this answer









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        3





        $begingroup$

        Caution:



        $$(x+1)^2=9implies x+1=color{red}{pm 3}impliesbegin{cases}x_1+1=-3implies x_1=-4\{}\x_2+1=3implies x_2=2end{cases}$$






        share|cite|improve this answer









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        Caution:



        $$(x+1)^2=9implies x+1=color{red}{pm 3}impliesbegin{cases}x_1+1=-3implies x_1=-4\{}\x_2+1=3implies x_2=2end{cases}$$







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        answered Jan 17 at 21:13









        DonAntonioDonAntonio

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