Mixing
Solve second degree polynomial by “completing the square”
$begingroup$
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
edited Jan 17 at 21:13
Arturo Magidin
264k34587915
asked Jan 17 at 21:10
Ryan CameronRyan Cameron
697
$endgroup$
add a comment |
$begingroup$
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
edited Jan 17 at 21:13
Arturo Magidin
264k34587915
asked Jan 17 at 21:10
Ryan CameronRyan Cameron
697
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
edited Jan 17 at 21:13
Arturo Magidin
264k34587915
asked Jan 17 at 21:10
Ryan CameronRyan Cameron
697
$endgroup$
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
edited Jan 17 at 21:13
Arturo Magidin
264k34587915
asked Jan 17 at 21:10
Ryan CameronRyan Cameron
697
edited Jan 17 at 21:13
Arturo Magidin
264k34587915
asked Jan 17 at 21:10
Ryan CameronRyan Cameron
697
edited Jan 17 at 21:13
Arturo Magidin
264k34587915
edited Jan 17 at 21:13
Arturo Magidin
264k34587915
edited Jan 17 at 21:13
Arturo Magidin
264k34587915
264k34587915
asked Jan 17 at 21:10
Ryan CameronRyan Cameron
697
asked Jan 17 at 21:10
Ryan CameronRyan Cameron
697
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
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$begingroup$
Caution:
$$(x+1)^2=9implies x+1=color{red}{pm 3}impliesbegin{cases}x_1+1=-3implies x_1=-4\{}\x_2+1=3implies x_2=2end{cases}$$
answered Jan 17 at 21:13
DonAntonioDonAntonio
179k1494230
$endgroup$
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
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oldest
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$begingroup$
Caution:
$$(x+1)^2=9implies x+1=color{red}{pm 3}impliesbegin{cases}x_1+1=-3implies x_1=-4\{}\x_2+1=3implies x_2=2end{cases}$$
answered Jan 17 at 21:13
DonAntonioDonAntonio
179k1494230
$endgroup$
add a comment |
$begingroup$
Caution:
$$(x+1)^2=9implies x+1=color{red}{pm 3}impliesbegin{cases}x_1+1=-3implies x_1=-4\{}\x_2+1=3implies x_2=2end{cases}$$
answered Jan 17 at 21:13
DonAntonioDonAntonio
179k1494230
$endgroup$
add a comment |
$begingroup$
Caution:
$$(x+1)^2=9implies x+1=color{red}{pm 3}impliesbegin{cases}x_1+1=-3implies x_1=-4\{}\x_2+1=3implies x_2=2end{cases}$$
answered Jan 17 at 21:13
DonAntonioDonAntonio
179k1494230
$endgroup$
Caution:
$$(x+1)^2=9implies x+1=color{red}{pm 3}impliesbegin{cases}x_1+1=-3implies x_1=-4\{}\x_2+1=3implies x_2=2end{cases}$$
answered Jan 17 at 21:13
DonAntonioDonAntonio
179k1494230
answered Jan 17 at 21:13
DonAntonioDonAntonio
179k1494230
answered Jan 17 at 21:13
DonAntonioDonAntonio
179k1494230
answered Jan 17 at 21:13
DonAntonioDonAntonio
179k1494230
179k1494230
add a comment |
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$begingroup$
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