Mixing
How does this equation always work? $sqrt{x times (x+2)+1} = x + 1$ for non-negative $x$, and $=|x|-1$ for...
$begingroup$
When I was playing with new calculator's functions, somehow I managed to get a formula, which works with all real numbers (negative number have a slight change). I had asked my teacher about it, but never figured out why it works:
Non-Negative:
$$forall x in Bbb R_0^+: sqrt{x times (x+2)+1} = x + 1$$
Negative:
$$forall x in Bbb R^-: sqrt{x times (x+2)+1} = |x| - 1$$
algebra-precalculus
edited Jan 2 at 22:48
Blue
47.8k870152
asked Jan 2 at 22:25
Dia. FrostDia. Frost
213
$endgroup$
add a comment |
$begingroup$
When I was playing with new calculator's functions, somehow I managed to get a formula, which works with all real numbers (negative number have a slight change). I had asked my teacher about it, but never figured out why it works:
Non-Negative:
$$forall x in Bbb R_0^+: sqrt{x times (x+2)+1} = x + 1$$
Negative:
$$forall x in Bbb R^-: sqrt{x times (x+2)+1} = |x| - 1$$
algebra-precalculus
edited Jan 2 at 22:48
Blue
47.8k870152
asked Jan 2 at 22:25
Dia. FrostDia. Frost
213
$endgroup$
2
$begingroup$
$x(x+2)+1=x^2+2x+1=(x+1)^2$.
$endgroup$
– A.Γ.
Jan 2 at 22:27
1
$begingroup$
Your second formula is not true for $x=-1/2$.
$endgroup$
– farruhota
Jan 2 at 22:34
$begingroup$
Check your edited second equation. Are you sure it works for $-1 < x < 0$?
$endgroup$
– Deepak
Jan 2 at 22:37
add a comment |
$begingroup$
When I was playing with new calculator's functions, somehow I managed to get a formula, which works with all real numbers (negative number have a slight change). I had asked my teacher about it, but never figured out why it works:
Non-Negative:
$$forall x in Bbb R_0^+: sqrt{x times (x+2)+1} = x + 1$$
Negative:
$$forall x in Bbb R^-: sqrt{x times (x+2)+1} = |x| - 1$$
algebra-precalculus
edited Jan 2 at 22:48
Blue
47.8k870152
asked Jan 2 at 22:25
Dia. FrostDia. Frost
213
$endgroup$
When I was playing with new calculator's functions, somehow I managed to get a formula, which works with all real numbers (negative number have a slight change). I had asked my teacher about it, but never figured out why it works:
Non-Negative:
$$forall x in Bbb R_0^+: sqrt{x times (x+2)+1} = x + 1$$
Negative:
$$forall x in Bbb R^-: sqrt{x times (x+2)+1} = |x| - 1$$
algebra-precalculus
algebra-precalculus
edited Jan 2 at 22:48
Blue
47.8k870152
asked Jan 2 at 22:25
Dia. FrostDia. Frost
213
edited Jan 2 at 22:48
Blue
47.8k870152
asked Jan 2 at 22:25
Dia. FrostDia. Frost
213
edited Jan 2 at 22:48
Blue
47.8k870152
edited Jan 2 at 22:48
Blue
47.8k870152
edited Jan 2 at 22:48
Blue
47.8k870152
47.8k870152
asked Jan 2 at 22:25
Dia. FrostDia. Frost
213
asked Jan 2 at 22:25
Dia. FrostDia. Frost
213
asked Jan 2 at 22:25
Dia. FrostDia. Frost
213
213
2
$begingroup$
$x(x+2)+1=x^2+2x+1=(x+1)^2$.
$endgroup$
– A.Γ.
Jan 2 at 22:27
1
$begingroup$
Your second formula is not true for $x=-1/2$.
$endgroup$
– farruhota
Jan 2 at 22:34
$begingroup$
Check your edited second equation. Are you sure it works for $-1 < x < 0$?
$endgroup$
– Deepak
Jan 2 at 22:37
add a comment |
2
$begingroup$
$x(x+2)+1=x^2+2x+1=(x+1)^2$.
$endgroup$
– A.Γ.
Jan 2 at 22:27
1
$begingroup$
Your second formula is not true for $x=-1/2$.
$endgroup$
– farruhota
Jan 2 at 22:34
$begingroup$
Check your edited second equation. Are you sure it works for $-1 < x < 0$?
$endgroup$
– Deepak
Jan 2 at 22:37
2
2
$begingroup$
$x(x+2)+1=x^2+2x+1=(x+1)^2$.
$endgroup$
– A.Γ.
Jan 2 at 22:27
$begingroup$
$x(x+2)+1=x^2+2x+1=(x+1)^2$.
$endgroup$
– A.Γ.
Jan 2 at 22:27
1
1
$begingroup$
Your second formula is not true for $x=-1/2$.
$endgroup$
– farruhota
Jan 2 at 22:34
$begingroup$
Your second formula is not true for $x=-1/2$.
$endgroup$
– farruhota
Jan 2 at 22:34
$begingroup$
Check your edited second equation. Are you sure it works for $-1 < x < 0$?
$endgroup$
– Deepak
Jan 2 at 22:37
$begingroup$
Check your edited second equation. Are you sure it works for $-1 < x < 0$?
$endgroup$
– Deepak
Jan 2 at 22:37
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Remark that $x cdot (x+2) + 1 = x^2 + 2x + 1 = (x+1)^2$, so:
$$
sqrt{x cdot (x+2) + 1} = sqrt{(x+1)^2} = |x+1|
$$
You may also note that:
- For $x geq -1$, we have $|x+1| = x+1$
- For $x leq -1$, we have $|x+1| = -(x+1) = -x-1 = |x|-1$ (since $x<0$)
This shows why your formula works in the ranges it does. Note, though, that your ranges aren't quite right: for $x$ in $[-1,0]$, your first formula applies (try for example with $x=-1/2$: you get $1/2$, not $-1/2$).
answered Jan 2 at 22:35
SamboSambo
2,1822532
$endgroup$
$begingroup$
Would you mind, clarifying how I would denote the first formula with all real numbers from -0.999 (periodically) to positive infinity?
$endgroup$
– Dia. Frost
Jan 2 at 22:42
$begingroup$
Do you mean how to denote that range, i.e. as opposed to $mathbb{R}^+_0$?
$endgroup$
– Sambo
Jan 2 at 23:51
1
$begingroup$
Also, I hope you know that $-0.9999$ (periodically) equals (exactly) $-1$, and what you said is NOT how to denote the range $(-1,infty)$.
$endgroup$
– Sambo
Jan 2 at 23:52
add a comment |
$begingroup$
The right calculation is the follows:$$sqrt{(x+1)^2}=|x+1|$$
answered Jan 2 at 22:27
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.7k42864
$endgroup$
add a comment |
$begingroup$
Note that you have a perfect square trinomial.
$$x(x+2)+1 = x^2+2x+1$$
You probably know that by binomial expansion, $(apm b)^2 = a^2pm 2ab+b^2$.
$$(apm b)^2 = (apm b)(apm b) = a^2pm abpm ba +b^2 = a^2pm2ab+b^2$$
Notice the trinomial $x^2+2x+1$. You can see there are two perfect squares: $x^2$ and $1$. Also, the middle term is twice the product of their squares. Hence, the trinomial is of the form $a^2+2ab+b^2$. Factoring, you get
$$x^2+2(x)(1)+1^2 = (x+1)^2$$
So you have$sqrt{(x+1)^2}$. You also have $sqrt{a^2} = vert avert$, so
$$sqrt{(x+1)^2} = vert x+1vert = begin{cases} x+1; quad x geq -1 \ -x-1 = vert xvert-1; quad x < -1end{cases}$$
answered Jan 2 at 22:43
KM101KM101
5,9861523
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Remark that $x cdot (x+2) + 1 = x^2 + 2x + 1 = (x+1)^2$, so:
$$
sqrt{x cdot (x+2) + 1} = sqrt{(x+1)^2} = |x+1|
$$
You may also note that:
- For $x geq -1$, we have $|x+1| = x+1$
- For $x leq -1$, we have $|x+1| = -(x+1) = -x-1 = |x|-1$ (since $x<0$)
This shows why your formula works in the ranges it does. Note, though, that your ranges aren't quite right: for $x$ in $[-1,0]$, your first formula applies (try for example with $x=-1/2$: you get $1/2$, not $-1/2$).
answered Jan 2 at 22:35
SamboSambo
2,1822532
$endgroup$
$begingroup$
Would you mind, clarifying how I would denote the first formula with all real numbers from -0.999 (periodically) to positive infinity?
$endgroup$
– Dia. Frost
Jan 2 at 22:42
$begingroup$
Do you mean how to denote that range, i.e. as opposed to $mathbb{R}^+_0$?
$endgroup$
– Sambo
Jan 2 at 23:51
1
$begingroup$
Also, I hope you know that $-0.9999$ (periodically) equals (exactly) $-1$, and what you said is NOT how to denote the range $(-1,infty)$.
$endgroup$
– Sambo
Jan 2 at 23:52
add a comment |
$begingroup$
Remark that $x cdot (x+2) + 1 = x^2 + 2x + 1 = (x+1)^2$, so:
$$
sqrt{x cdot (x+2) + 1} = sqrt{(x+1)^2} = |x+1|
$$
You may also note that:
- For $x geq -1$, we have $|x+1| = x+1$
- For $x leq -1$, we have $|x+1| = -(x+1) = -x-1 = |x|-1$ (since $x<0$)
This shows why your formula works in the ranges it does. Note, though, that your ranges aren't quite right: for $x$ in $[-1,0]$, your first formula applies (try for example with $x=-1/2$: you get $1/2$, not $-1/2$).
answered Jan 2 at 22:35
SamboSambo
2,1822532
$endgroup$
$begingroup$
Would you mind, clarifying how I would denote the first formula with all real numbers from -0.999 (periodically) to positive infinity?
$endgroup$
– Dia. Frost
Jan 2 at 22:42
$begingroup$
Do you mean how to denote that range, i.e. as opposed to $mathbb{R}^+_0$?
$endgroup$
– Sambo
Jan 2 at 23:51
1
$begingroup$
Also, I hope you know that $-0.9999$ (periodically) equals (exactly) $-1$, and what you said is NOT how to denote the range $(-1,infty)$.
$endgroup$
– Sambo
Jan 2 at 23:52
add a comment |
$begingroup$
Remark that $x cdot (x+2) + 1 = x^2 + 2x + 1 = (x+1)^2$, so:
$$
sqrt{x cdot (x+2) + 1} = sqrt{(x+1)^2} = |x+1|
$$
You may also note that:
- For $x geq -1$, we have $|x+1| = x+1$
- For $x leq -1$, we have $|x+1| = -(x+1) = -x-1 = |x|-1$ (since $x<0$)
This shows why your formula works in the ranges it does. Note, though, that your ranges aren't quite right: for $x$ in $[-1,0]$, your first formula applies (try for example with $x=-1/2$: you get $1/2$, not $-1/2$).
answered Jan 2 at 22:35
SamboSambo
2,1822532
$endgroup$
Remark that $x cdot (x+2) + 1 = x^2 + 2x + 1 = (x+1)^2$, so:
$$
sqrt{x cdot (x+2) + 1} = sqrt{(x+1)^2} = |x+1|
$$
You may also note that:
- For $x geq -1$, we have $|x+1| = x+1$
- For $x leq -1$, we have $|x+1| = -(x+1) = -x-1 = |x|-1$ (since $x<0$)
This shows why your formula works in the ranges it does. Note, though, that your ranges aren't quite right: for $x$ in $[-1,0]$, your first formula applies (try for example with $x=-1/2$: you get $1/2$, not $-1/2$).
answered Jan 2 at 22:35
SamboSambo
2,1822532
answered Jan 2 at 22:35
SamboSambo
2,1822532
answered Jan 2 at 22:35
SamboSambo
2,1822532
answered Jan 2 at 22:35
SamboSambo
2,1822532
2,1822532
$begingroup$
Would you mind, clarifying how I would denote the first formula with all real numbers from -0.999 (periodically) to positive infinity?
$endgroup$
– Dia. Frost
Jan 2 at 22:42
$begingroup$
Do you mean how to denote that range, i.e. as opposed to $mathbb{R}^+_0$?
$endgroup$
– Sambo
Jan 2 at 23:51
1
$begingroup$
Also, I hope you know that $-0.9999$ (periodically) equals (exactly) $-1$, and what you said is NOT how to denote the range $(-1,infty)$.
$endgroup$
– Sambo
Jan 2 at 23:52
add a comment |
$begingroup$
Would you mind, clarifying how I would denote the first formula with all real numbers from -0.999 (periodically) to positive infinity?
$endgroup$
– Dia. Frost
Jan 2 at 22:42
$begingroup$
Do you mean how to denote that range, i.e. as opposed to $mathbb{R}^+_0$?
$endgroup$
– Sambo
Jan 2 at 23:51
1
$begingroup$
Also, I hope you know that $-0.9999$ (periodically) equals (exactly) $-1$, and what you said is NOT how to denote the range $(-1,infty)$.
$endgroup$
– Sambo
Jan 2 at 23:52
$begingroup$
Would you mind, clarifying how I would denote the first formula with all real numbers from -0.999 (periodically) to positive infinity?
$endgroup$
– Dia. Frost
Jan 2 at 22:42
$begingroup$
Would you mind, clarifying how I would denote the first formula with all real numbers from -0.999 (periodically) to positive infinity?
$endgroup$
– Dia. Frost
Jan 2 at 22:42
$begingroup$
Do you mean how to denote that range, i.e. as opposed to $mathbb{R}^+_0$?
$endgroup$
– Sambo
Jan 2 at 23:51
$begingroup$
Do you mean how to denote that range, i.e. as opposed to $mathbb{R}^+_0$?
$endgroup$
– Sambo
Jan 2 at 23:51
1
1
$begingroup$
Also, I hope you know that $-0.9999$ (periodically) equals (exactly) $-1$, and what you said is NOT how to denote the range $(-1,infty)$.
$endgroup$
– Sambo
Jan 2 at 23:52
$begingroup$
Also, I hope you know that $-0.9999$ (periodically) equals (exactly) $-1$, and what you said is NOT how to denote the range $(-1,infty)$.
$endgroup$
– Sambo
Jan 2 at 23:52
add a comment |
$begingroup$
The right calculation is the follows:$$sqrt{(x+1)^2}=|x+1|$$
answered Jan 2 at 22:27
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.7k42864
$endgroup$
add a comment |
$begingroup$
The right calculation is the follows:$$sqrt{(x+1)^2}=|x+1|$$
answered Jan 2 at 22:27
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.7k42864
$endgroup$
add a comment |
$begingroup$
The right calculation is the follows:$$sqrt{(x+1)^2}=|x+1|$$
answered Jan 2 at 22:27
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.7k42864
$endgroup$
The right calculation is the follows:$$sqrt{(x+1)^2}=|x+1|$$
answered Jan 2 at 22:27
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.7k42864
answered Jan 2 at 22:27
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.7k42864
answered Jan 2 at 22:27
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.7k42864
answered Jan 2 at 22:27
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.7k42864
73.7k42864
add a comment |
add a comment |
$begingroup$
Note that you have a perfect square trinomial.
$$x(x+2)+1 = x^2+2x+1$$
You probably know that by binomial expansion, $(apm b)^2 = a^2pm 2ab+b^2$.
$$(apm b)^2 = (apm b)(apm b) = a^2pm abpm ba +b^2 = a^2pm2ab+b^2$$
Notice the trinomial $x^2+2x+1$. You can see there are two perfect squares: $x^2$ and $1$. Also, the middle term is twice the product of their squares. Hence, the trinomial is of the form $a^2+2ab+b^2$. Factoring, you get
$$x^2+2(x)(1)+1^2 = (x+1)^2$$
So you have$sqrt{(x+1)^2}$. You also have $sqrt{a^2} = vert avert$, so
$$sqrt{(x+1)^2} = vert x+1vert = begin{cases} x+1; quad x geq -1 \ -x-1 = vert xvert-1; quad x < -1end{cases}$$
answered Jan 2 at 22:43
KM101KM101
5,9861523
$endgroup$
add a comment |
$begingroup$
Note that you have a perfect square trinomial.
$$x(x+2)+1 = x^2+2x+1$$
You probably know that by binomial expansion, $(apm b)^2 = a^2pm 2ab+b^2$.
$$(apm b)^2 = (apm b)(apm b) = a^2pm abpm ba +b^2 = a^2pm2ab+b^2$$
Notice the trinomial $x^2+2x+1$. You can see there are two perfect squares: $x^2$ and $1$. Also, the middle term is twice the product of their squares. Hence, the trinomial is of the form $a^2+2ab+b^2$. Factoring, you get
$$x^2+2(x)(1)+1^2 = (x+1)^2$$
So you have$sqrt{(x+1)^2}$. You also have $sqrt{a^2} = vert avert$, so
$$sqrt{(x+1)^2} = vert x+1vert = begin{cases} x+1; quad x geq -1 \ -x-1 = vert xvert-1; quad x < -1end{cases}$$
answered Jan 2 at 22:43
KM101KM101
5,9861523
$endgroup$
add a comment |
$begingroup$
Note that you have a perfect square trinomial.
$$x(x+2)+1 = x^2+2x+1$$
You probably know that by binomial expansion, $(apm b)^2 = a^2pm 2ab+b^2$.
$$(apm b)^2 = (apm b)(apm b) = a^2pm abpm ba +b^2 = a^2pm2ab+b^2$$
Notice the trinomial $x^2+2x+1$. You can see there are two perfect squares: $x^2$ and $1$. Also, the middle term is twice the product of their squares. Hence, the trinomial is of the form $a^2+2ab+b^2$. Factoring, you get
$$x^2+2(x)(1)+1^2 = (x+1)^2$$
So you have$sqrt{(x+1)^2}$. You also have $sqrt{a^2} = vert avert$, so
$$sqrt{(x+1)^2} = vert x+1vert = begin{cases} x+1; quad x geq -1 \ -x-1 = vert xvert-1; quad x < -1end{cases}$$
answered Jan 2 at 22:43
KM101KM101
5,9861523
$endgroup$
Note that you have a perfect square trinomial.
$$x(x+2)+1 = x^2+2x+1$$
You probably know that by binomial expansion, $(apm b)^2 = a^2pm 2ab+b^2$.
$$(apm b)^2 = (apm b)(apm b) = a^2pm abpm ba +b^2 = a^2pm2ab+b^2$$
Notice the trinomial $x^2+2x+1$. You can see there are two perfect squares: $x^2$ and $1$. Also, the middle term is twice the product of their squares. Hence, the trinomial is of the form $a^2+2ab+b^2$. Factoring, you get
$$x^2+2(x)(1)+1^2 = (x+1)^2$$
So you have$sqrt{(x+1)^2}$. You also have $sqrt{a^2} = vert avert$, so
$$sqrt{(x+1)^2} = vert x+1vert = begin{cases} x+1; quad x geq -1 \ -x-1 = vert xvert-1; quad x < -1end{cases}$$
answered Jan 2 at 22:43
KM101KM101
5,9861523
answered Jan 2 at 22:43
KM101KM101
5,9861523
answered Jan 2 at 22:43
KM101KM101
5,9861523
answered Jan 2 at 22:43
KM101KM101
5,9861523
5,9861523
add a comment |
add a comment |
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2
$begingroup$
$x(x+2)+1=x^2+2x+1=(x+1)^2$.
$endgroup$
– A.Γ.
Jan 2 at 22:27
1
$begingroup$
Your second formula is not true for $x=-1/2$.
$endgroup$
– farruhota
Jan 2 at 22:34
$begingroup$
Check your edited second equation. Are you sure it works for $-1 < x < 0$?
$endgroup$
– Deepak
Jan 2 at 22:37