Mixing
How to “Re-write completing the square”: $x^2+x+1$
$begingroup$
The exercise asks to "Re-write completing the square": $$x^2+x+1$$
The answer is: $$left(x+frac{1}{2}right)^2+frac{3}{4}$$
I don't even understand what it means with "Re-write completing the square"..
What's the steps to solve this?
algebra-precalculus polynomials quadratics completing-the-square
edited Feb 3 at 9:06
Martin Sleziak
45k10123277
asked Feb 5 '11 at 21:46
Tom BritoTom Brito
3861311
$endgroup$
add a comment |
$begingroup$
The exercise asks to "Re-write completing the square": $$x^2+x+1$$
The answer is: $$left(x+frac{1}{2}right)^2+frac{3}{4}$$
I don't even understand what it means with "Re-write completing the square"..
What's the steps to solve this?
algebra-precalculus polynomials quadratics completing-the-square
edited Feb 3 at 9:06
Martin Sleziak
45k10123277
asked Feb 5 '11 at 21:46
Tom BritoTom Brito
3861311
$endgroup$
4
$begingroup$
en.wikipedia.org/wiki/Completing_the_square
$endgroup$
– Dinesh
Feb 5 '11 at 21:48
add a comment |
$begingroup$
The exercise asks to "Re-write completing the square": $$x^2+x+1$$
The answer is: $$left(x+frac{1}{2}right)^2+frac{3}{4}$$
I don't even understand what it means with "Re-write completing the square"..
What's the steps to solve this?
algebra-precalculus polynomials quadratics completing-the-square
edited Feb 3 at 9:06
Martin Sleziak
45k10123277
asked Feb 5 '11 at 21:46
Tom BritoTom Brito
3861311
$endgroup$
The exercise asks to "Re-write completing the square": $$x^2+x+1$$
The answer is: $$left(x+frac{1}{2}right)^2+frac{3}{4}$$
I don't even understand what it means with "Re-write completing the square"..
What's the steps to solve this?
algebra-precalculus polynomials quadratics completing-the-square
algebra-precalculus polynomials quadratics completing-the-square
edited Feb 3 at 9:06
Martin Sleziak
45k10123277
asked Feb 5 '11 at 21:46
Tom BritoTom Brito
3861311
edited Feb 3 at 9:06
Martin Sleziak
45k10123277
asked Feb 5 '11 at 21:46
Tom BritoTom Brito
3861311
edited Feb 3 at 9:06
Martin Sleziak
45k10123277
edited Feb 3 at 9:06
Martin Sleziak
45k10123277
edited Feb 3 at 9:06
Martin Sleziak
45k10123277
45k10123277
asked Feb 5 '11 at 21:46
Tom BritoTom Brito
3861311
asked Feb 5 '11 at 21:46
Tom BritoTom Brito
3861311
asked Feb 5 '11 at 21:46
Tom BritoTom Brito
3861311
3861311
4
$begingroup$
en.wikipedia.org/wiki/Completing_the_square
$endgroup$
– Dinesh
Feb 5 '11 at 21:48
add a comment |
4
$begingroup$
en.wikipedia.org/wiki/Completing_the_square
$endgroup$
– Dinesh
Feb 5 '11 at 21:48
4
4
$begingroup$
en.wikipedia.org/wiki/Completing_the_square
$endgroup$
– Dinesh
Feb 5 '11 at 21:48
$begingroup$
en.wikipedia.org/wiki/Completing_the_square
$endgroup$
– Dinesh
Feb 5 '11 at 21:48
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
Remember the formula for the square of a binomial:
$$(a+b)^2 = a^2 + 2ab + b^2.$$
Now, when you see $x^2+x+1$, you want to think of $x^2+x$ as the first two terms you get in expanding the binomial $(x+c)^2$ for some $c$; that is,
$$x^2 + x + cdots = (x+c)^2.$$
Since the middle term should be $2cx$, and you have $x$, that means that you want $2c=1$, or $c=frac{1}{2}$.
But if you have $(x+frac{1}{2})^2$, you get $x^2 + x + frac{1}{4}$. Since all you have is $x^2+x$, you complete the square by adding the missing "$frac{1}{4}$". Since you are not allowed to just add constants willy-nilly, you must also cancel it out by subtracting $frac{1}{4}$. So:
begin{align*}
x^2 + x + 2 &= (x^2 + x + cdots) + 1\
&=left( x^2 + 2left(frac{1}{2}right)x + cdots right) + 1 &&mbox{figuring out what $c$ is}\
&= left(x^2 + 2left(frac{1}{2}right)x + left(frac{1}{2}right)^2right) -frac{1}{4} + 1 &&mbox{completing the square}\
&= left(x+frac{1}{2}right)^2 + frac{3}{4}.
end{align*}
answered Feb 5 '11 at 21:59
Arturo MagidinArturo Magidin
266k34591922
$endgroup$
1
$begingroup$
Looking at wikipedia, it says that the form is $a(x-h)^2+k$, so this question asked for complete the square of $2x^2+x+1$ the answer would be $2(x+frac{1}{2})^2+frac{3}{4}$ correct?
$endgroup$
– Tom Brito
Feb 6 '11 at 0:36
1
$begingroup$
@Tom: If you multiply it out, you'll see it doesn't work out right (and you should have done that before asking!). To get it right: First factor out the $2$ from the $x^2$ and $x$ terms to get $2(x^2+frac{1}{2}x) + 1$. Then complete the square as above: here $2c=frac{1}{2}$, so $c=frac{1}{4}$, so you need to add $c^2=frac{1}{16}$; since it is multiplied by the $2$, you are really adding $frac{2}{16}=frac{1}{8}$, which then is subtracted. So you get$$2x^2+x+1 = 2(x^2+frac{1}{2}x) + 1 = 2(x^2+frac{1}{2}x+frac{1}{16}) + 1-frac{1}{8} = 2(x+frac{1}{4})^2 + frac{7}{8}.$$
$endgroup$
– Arturo Magidin
Feb 6 '11 at 0:44
add a comment |
$begingroup$
"Completing the square" is a standard step in solving a quadratic equation. To see how it helps, consider the following: The general formulation of a quadratic equation is $ax^2+bx+c = 0$ with $a neq 0$. Let us say we are tasked with solving this equation, i.e., finding values of $x$ that satisfy this equation. To start with, note that this equation is easy to solve if $b=0$. Then the equation looks like $ax^2 + c = 0$ which would simplify to $x = pm sqrt{frac{-c}{a}}$.
"Completing the square" is a step that takes a general quadratic and reduces it to the form of the simple quadratic above. It does this with a substitution $y = x + frac{b}{2a}$. Then, $ay^2 = ax^2 + bx + frac{b^2}{4a}$ which means the general quadratic can be written as $ay^2-frac{b^2}{4a} + c = 0$ or equivalently as
$$ a(x+frac{b}{2a})^2 = frac{b^2}{4a} - c $$
This equation is easy to solve and yields the two roots of the general quadratic equation. Note that the above equation is equivalent to the one we started with ($ax^2+bx+c = 0$) for the purposes of finding the roots. This process of rewriting is called completing the square. This is the point behind rewriting $x^2+x+1$ as $(x+frac{1}{2})^2 + frac{3}{4}$.
answered Feb 5 '11 at 21:58
DineshDinesh
2,2391923
$endgroup$
$begingroup$
"eliminating the middle term" is a bit misleading. One hasn't eliminated it. Rather one has "absorbed" it into the the square term by "completing" the "partial square" comprised of the two highest degree terms.
$endgroup$
– Bill Dubuque
Feb 5 '11 at 22:08
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@Bill: I agree. I will edit the answer accordingly. Thanks.
$endgroup$
– Dinesh
Feb 5 '11 at 22:16
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Arturo's answer explains the reason behind the name "completing the square" part very well. I will leave this answer up for the derivation of the roots of the quadratic part.
$endgroup$
– Dinesh
Feb 5 '11 at 22:22
add a comment |
$begingroup$
One can rewrite a degree $rm:n > 1:$ polynomial $rm f(x) = x^n + b x^{n-1} + cdots $ into a form such that its two highest degree terms are "absorbed" into a perfect $rm: n$'th power of a linear polynomial, namely
$rmquadquad f(x) = (x + b/n)^n - g(x) $ where $rm g(x) = (x+b/n)^n - f(x) $ has degree $rm:le: n-2$
When $rm n = 2 $ this is called completing the square - esp. when used to solve a quadratic equation.$ $ If $rm g(x) = g $ is constant (as is always true when $rm n = 2:$) then this yields a closed form for the roots of $rm: f(x):, $ namely $rm x = sqrt[n]{g}-b/2:.$
answered Feb 5 '11 at 22:27
Bill DubuqueBill Dubuque
214k29197659
$endgroup$
add a comment |
$begingroup$
See this video and
its sequel to see the process worked in real time.
answered Feb 6 '11 at 16:06
Scott CarterScott Carter
2,2411113
$endgroup$
add a comment |
$begingroup$
As Arturo points out what you have to observe is the coefficient.
begin{align*}
x^{2}+x+1 & = x^{2} + x + frac{1+3}{4} \ & = x^{2} + x + frac{1}{4} + frac{3}{4} \ & = Bigl(x+frac{1}{2}Bigr)^{2} + frac{3}{4}
end{align*}
I am sure once you get used to such type of things you shall not have trouble in doing such problems. Solve more problems based on this type. Suppose you have the coefficient of $x$ as $a$ note that $a^{2}/4$ should be added and subtracted from the constant term. What i mean by this is: Suppose you have something of this type $x^{2}+ax + b^{2}$ then you can write this as $(x+a/2)^{2} + b^{2}-a^{2}/{4}$.
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Remember the formula for the square of a binomial:
$$(a+b)^2 = a^2 + 2ab + b^2.$$
Now, when you see $x^2+x+1$, you want to think of $x^2+x$ as the first two terms you get in expanding the binomial $(x+c)^2$ for some $c$; that is,
$$x^2 + x + cdots = (x+c)^2.$$
Since the middle term should be $2cx$, and you have $x$, that means that you want $2c=1$, or $c=frac{1}{2}$.
But if you have $(x+frac{1}{2})^2$, you get $x^2 + x + frac{1}{4}$. Since all you have is $x^2+x$, you complete the square by adding the missing "$frac{1}{4}$". Since you are not allowed to just add constants willy-nilly, you must also cancel it out by subtracting $frac{1}{4}$. So:
begin{align*}
x^2 + x + 2 &= (x^2 + x + cdots) + 1\
&=left( x^2 + 2left(frac{1}{2}right)x + cdots right) + 1 &&mbox{figuring out what $c$ is}\
&= left(x^2 + 2left(frac{1}{2}right)x + left(frac{1}{2}right)^2right) -frac{1}{4} + 1 &&mbox{completing the square}\
&= left(x+frac{1}{2}right)^2 + frac{3}{4}.
end{align*}
answered Feb 5 '11 at 21:59
Arturo MagidinArturo Magidin
266k34591922
$endgroup$
1
$begingroup$
Looking at wikipedia, it says that the form is $a(x-h)^2+k$, so this question asked for complete the square of $2x^2+x+1$ the answer would be $2(x+frac{1}{2})^2+frac{3}{4}$ correct?
$endgroup$
– Tom Brito
Feb 6 '11 at 0:36
1
$begingroup$
@Tom: If you multiply it out, you'll see it doesn't work out right (and you should have done that before asking!). To get it right: First factor out the $2$ from the $x^2$ and $x$ terms to get $2(x^2+frac{1}{2}x) + 1$. Then complete the square as above: here $2c=frac{1}{2}$, so $c=frac{1}{4}$, so you need to add $c^2=frac{1}{16}$; since it is multiplied by the $2$, you are really adding $frac{2}{16}=frac{1}{8}$, which then is subtracted. So you get$$2x^2+x+1 = 2(x^2+frac{1}{2}x) + 1 = 2(x^2+frac{1}{2}x+frac{1}{16}) + 1-frac{1}{8} = 2(x+frac{1}{4})^2 + frac{7}{8}.$$
$endgroup$
– Arturo Magidin
Feb 6 '11 at 0:44
add a comment |
$begingroup$
Remember the formula for the square of a binomial:
$$(a+b)^2 = a^2 + 2ab + b^2.$$
Now, when you see $x^2+x+1$, you want to think of $x^2+x$ as the first two terms you get in expanding the binomial $(x+c)^2$ for some $c$; that is,
$$x^2 + x + cdots = (x+c)^2.$$
Since the middle term should be $2cx$, and you have $x$, that means that you want $2c=1$, or $c=frac{1}{2}$.
But if you have $(x+frac{1}{2})^2$, you get $x^2 + x + frac{1}{4}$. Since all you have is $x^2+x$, you complete the square by adding the missing "$frac{1}{4}$". Since you are not allowed to just add constants willy-nilly, you must also cancel it out by subtracting $frac{1}{4}$. So:
begin{align*}
x^2 + x + 2 &= (x^2 + x + cdots) + 1\
&=left( x^2 + 2left(frac{1}{2}right)x + cdots right) + 1 &&mbox{figuring out what $c$ is}\
&= left(x^2 + 2left(frac{1}{2}right)x + left(frac{1}{2}right)^2right) -frac{1}{4} + 1 &&mbox{completing the square}\
&= left(x+frac{1}{2}right)^2 + frac{3}{4}.
end{align*}
answered Feb 5 '11 at 21:59
Arturo MagidinArturo Magidin
266k34591922
$endgroup$
1
$begingroup$
Looking at wikipedia, it says that the form is $a(x-h)^2+k$, so this question asked for complete the square of $2x^2+x+1$ the answer would be $2(x+frac{1}{2})^2+frac{3}{4}$ correct?
$endgroup$
– Tom Brito
Feb 6 '11 at 0:36
1
$begingroup$
@Tom: If you multiply it out, you'll see it doesn't work out right (and you should have done that before asking!). To get it right: First factor out the $2$ from the $x^2$ and $x$ terms to get $2(x^2+frac{1}{2}x) + 1$. Then complete the square as above: here $2c=frac{1}{2}$, so $c=frac{1}{4}$, so you need to add $c^2=frac{1}{16}$; since it is multiplied by the $2$, you are really adding $frac{2}{16}=frac{1}{8}$, which then is subtracted. So you get$$2x^2+x+1 = 2(x^2+frac{1}{2}x) + 1 = 2(x^2+frac{1}{2}x+frac{1}{16}) + 1-frac{1}{8} = 2(x+frac{1}{4})^2 + frac{7}{8}.$$
$endgroup$
– Arturo Magidin
Feb 6 '11 at 0:44
add a comment |
$begingroup$
Remember the formula for the square of a binomial:
$$(a+b)^2 = a^2 + 2ab + b^2.$$
Now, when you see $x^2+x+1$, you want to think of $x^2+x$ as the first two terms you get in expanding the binomial $(x+c)^2$ for some $c$; that is,
$$x^2 + x + cdots = (x+c)^2.$$
Since the middle term should be $2cx$, and you have $x$, that means that you want $2c=1$, or $c=frac{1}{2}$.
But if you have $(x+frac{1}{2})^2$, you get $x^2 + x + frac{1}{4}$. Since all you have is $x^2+x$, you complete the square by adding the missing "$frac{1}{4}$". Since you are not allowed to just add constants willy-nilly, you must also cancel it out by subtracting $frac{1}{4}$. So:
begin{align*}
x^2 + x + 2 &= (x^2 + x + cdots) + 1\
&=left( x^2 + 2left(frac{1}{2}right)x + cdots right) + 1 &&mbox{figuring out what $c$ is}\
&= left(x^2 + 2left(frac{1}{2}right)x + left(frac{1}{2}right)^2right) -frac{1}{4} + 1 &&mbox{completing the square}\
&= left(x+frac{1}{2}right)^2 + frac{3}{4}.
end{align*}
answered Feb 5 '11 at 21:59
Arturo MagidinArturo Magidin
266k34591922
$endgroup$
Remember the formula for the square of a binomial:
$$(a+b)^2 = a^2 + 2ab + b^2.$$
Now, when you see $x^2+x+1$, you want to think of $x^2+x$ as the first two terms you get in expanding the binomial $(x+c)^2$ for some $c$; that is,
$$x^2 + x + cdots = (x+c)^2.$$
Since the middle term should be $2cx$, and you have $x$, that means that you want $2c=1$, or $c=frac{1}{2}$.
But if you have $(x+frac{1}{2})^2$, you get $x^2 + x + frac{1}{4}$. Since all you have is $x^2+x$, you complete the square by adding the missing "$frac{1}{4}$". Since you are not allowed to just add constants willy-nilly, you must also cancel it out by subtracting $frac{1}{4}$. So:
begin{align*}
x^2 + x + 2 &= (x^2 + x + cdots) + 1\
&=left( x^2 + 2left(frac{1}{2}right)x + cdots right) + 1 &&mbox{figuring out what $c$ is}\
&= left(x^2 + 2left(frac{1}{2}right)x + left(frac{1}{2}right)^2right) -frac{1}{4} + 1 &&mbox{completing the square}\
&= left(x+frac{1}{2}right)^2 + frac{3}{4}.
end{align*}
answered Feb 5 '11 at 21:59
Arturo MagidinArturo Magidin
266k34591922
edited Feb 5 '11 at 22:10
answered Feb 5 '11 at 21:59
Arturo MagidinArturo Magidin
266k34591922
answered Feb 5 '11 at 21:59
Arturo MagidinArturo Magidin
266k34591922
answered Feb 5 '11 at 21:59
Arturo MagidinArturo Magidin
266k34591922
266k34591922
1
$begingroup$
Looking at wikipedia, it says that the form is $a(x-h)^2+k$, so this question asked for complete the square of $2x^2+x+1$ the answer would be $2(x+frac{1}{2})^2+frac{3}{4}$ correct?
$endgroup$
– Tom Brito
Feb 6 '11 at 0:36
1
$begingroup$
@Tom: If you multiply it out, you'll see it doesn't work out right (and you should have done that before asking!). To get it right: First factor out the $2$ from the $x^2$ and $x$ terms to get $2(x^2+frac{1}{2}x) + 1$. Then complete the square as above: here $2c=frac{1}{2}$, so $c=frac{1}{4}$, so you need to add $c^2=frac{1}{16}$; since it is multiplied by the $2$, you are really adding $frac{2}{16}=frac{1}{8}$, which then is subtracted. So you get$$2x^2+x+1 = 2(x^2+frac{1}{2}x) + 1 = 2(x^2+frac{1}{2}x+frac{1}{16}) + 1-frac{1}{8} = 2(x+frac{1}{4})^2 + frac{7}{8}.$$
$endgroup$
– Arturo Magidin
Feb 6 '11 at 0:44
add a comment |
1
$begingroup$
Looking at wikipedia, it says that the form is $a(x-h)^2+k$, so this question asked for complete the square of $2x^2+x+1$ the answer would be $2(x+frac{1}{2})^2+frac{3}{4}$ correct?
$endgroup$
– Tom Brito
Feb 6 '11 at 0:36
1
$begingroup$
@Tom: If you multiply it out, you'll see it doesn't work out right (and you should have done that before asking!). To get it right: First factor out the $2$ from the $x^2$ and $x$ terms to get $2(x^2+frac{1}{2}x) + 1$. Then complete the square as above: here $2c=frac{1}{2}$, so $c=frac{1}{4}$, so you need to add $c^2=frac{1}{16}$; since it is multiplied by the $2$, you are really adding $frac{2}{16}=frac{1}{8}$, which then is subtracted. So you get$$2x^2+x+1 = 2(x^2+frac{1}{2}x) + 1 = 2(x^2+frac{1}{2}x+frac{1}{16}) + 1-frac{1}{8} = 2(x+frac{1}{4})^2 + frac{7}{8}.$$
$endgroup$
– Arturo Magidin
Feb 6 '11 at 0:44
1
1
$begingroup$
Looking at wikipedia, it says that the form is $a(x-h)^2+k$, so this question asked for complete the square of $2x^2+x+1$ the answer would be $2(x+frac{1}{2})^2+frac{3}{4}$ correct?
$endgroup$
– Tom Brito
Feb 6 '11 at 0:36
$begingroup$
Looking at wikipedia, it says that the form is $a(x-h)^2+k$, so this question asked for complete the square of $2x^2+x+1$ the answer would be $2(x+frac{1}{2})^2+frac{3}{4}$ correct?
$endgroup$
– Tom Brito
Feb 6 '11 at 0:36
1
1
$begingroup$
@Tom: If you multiply it out, you'll see it doesn't work out right (and you should have done that before asking!). To get it right: First factor out the $2$ from the $x^2$ and $x$ terms to get $2(x^2+frac{1}{2}x) + 1$. Then complete the square as above: here $2c=frac{1}{2}$, so $c=frac{1}{4}$, so you need to add $c^2=frac{1}{16}$; since it is multiplied by the $2$, you are really adding $frac{2}{16}=frac{1}{8}$, which then is subtracted. So you get$$2x^2+x+1 = 2(x^2+frac{1}{2}x) + 1 = 2(x^2+frac{1}{2}x+frac{1}{16}) + 1-frac{1}{8} = 2(x+frac{1}{4})^2 + frac{7}{8}.$$
$endgroup$
– Arturo Magidin
Feb 6 '11 at 0:44
$begingroup$
@Tom: If you multiply it out, you'll see it doesn't work out right (and you should have done that before asking!). To get it right: First factor out the $2$ from the $x^2$ and $x$ terms to get $2(x^2+frac{1}{2}x) + 1$. Then complete the square as above: here $2c=frac{1}{2}$, so $c=frac{1}{4}$, so you need to add $c^2=frac{1}{16}$; since it is multiplied by the $2$, you are really adding $frac{2}{16}=frac{1}{8}$, which then is subtracted. So you get$$2x^2+x+1 = 2(x^2+frac{1}{2}x) + 1 = 2(x^2+frac{1}{2}x+frac{1}{16}) + 1-frac{1}{8} = 2(x+frac{1}{4})^2 + frac{7}{8}.$$
$endgroup$
– Arturo Magidin
Feb 6 '11 at 0:44
add a comment |
$begingroup$
"Completing the square" is a standard step in solving a quadratic equation. To see how it helps, consider the following: The general formulation of a quadratic equation is $ax^2+bx+c = 0$ with $a neq 0$. Let us say we are tasked with solving this equation, i.e., finding values of $x$ that satisfy this equation. To start with, note that this equation is easy to solve if $b=0$. Then the equation looks like $ax^2 + c = 0$ which would simplify to $x = pm sqrt{frac{-c}{a}}$.
"Completing the square" is a step that takes a general quadratic and reduces it to the form of the simple quadratic above. It does this with a substitution $y = x + frac{b}{2a}$. Then, $ay^2 = ax^2 + bx + frac{b^2}{4a}$ which means the general quadratic can be written as $ay^2-frac{b^2}{4a} + c = 0$ or equivalently as
$$ a(x+frac{b}{2a})^2 = frac{b^2}{4a} - c $$
This equation is easy to solve and yields the two roots of the general quadratic equation. Note that the above equation is equivalent to the one we started with ($ax^2+bx+c = 0$) for the purposes of finding the roots. This process of rewriting is called completing the square. This is the point behind rewriting $x^2+x+1$ as $(x+frac{1}{2})^2 + frac{3}{4}$.
answered Feb 5 '11 at 21:58
DineshDinesh
2,2391923
$endgroup$
$begingroup$
"eliminating the middle term" is a bit misleading. One hasn't eliminated it. Rather one has "absorbed" it into the the square term by "completing" the "partial square" comprised of the two highest degree terms.
$endgroup$
– Bill Dubuque
Feb 5 '11 at 22:08
$begingroup$
@Bill: I agree. I will edit the answer accordingly. Thanks.
$endgroup$
– Dinesh
Feb 5 '11 at 22:16
$begingroup$
Arturo's answer explains the reason behind the name "completing the square" part very well. I will leave this answer up for the derivation of the roots of the quadratic part.
$endgroup$
– Dinesh
Feb 5 '11 at 22:22
add a comment |
$begingroup$
"Completing the square" is a standard step in solving a quadratic equation. To see how it helps, consider the following: The general formulation of a quadratic equation is $ax^2+bx+c = 0$ with $a neq 0$. Let us say we are tasked with solving this equation, i.e., finding values of $x$ that satisfy this equation. To start with, note that this equation is easy to solve if $b=0$. Then the equation looks like $ax^2 + c = 0$ which would simplify to $x = pm sqrt{frac{-c}{a}}$.
"Completing the square" is a step that takes a general quadratic and reduces it to the form of the simple quadratic above. It does this with a substitution $y = x + frac{b}{2a}$. Then, $ay^2 = ax^2 + bx + frac{b^2}{4a}$ which means the general quadratic can be written as $ay^2-frac{b^2}{4a} + c = 0$ or equivalently as
$$ a(x+frac{b}{2a})^2 = frac{b^2}{4a} - c $$
This equation is easy to solve and yields the two roots of the general quadratic equation. Note that the above equation is equivalent to the one we started with ($ax^2+bx+c = 0$) for the purposes of finding the roots. This process of rewriting is called completing the square. This is the point behind rewriting $x^2+x+1$ as $(x+frac{1}{2})^2 + frac{3}{4}$.
answered Feb 5 '11 at 21:58
DineshDinesh
2,2391923
$endgroup$
$begingroup$
"eliminating the middle term" is a bit misleading. One hasn't eliminated it. Rather one has "absorbed" it into the the square term by "completing" the "partial square" comprised of the two highest degree terms.
$endgroup$
– Bill Dubuque
Feb 5 '11 at 22:08
$begingroup$
@Bill: I agree. I will edit the answer accordingly. Thanks.
$endgroup$
– Dinesh
Feb 5 '11 at 22:16
$begingroup$
Arturo's answer explains the reason behind the name "completing the square" part very well. I will leave this answer up for the derivation of the roots of the quadratic part.
$endgroup$
– Dinesh
Feb 5 '11 at 22:22
add a comment |
$begingroup$
"Completing the square" is a standard step in solving a quadratic equation. To see how it helps, consider the following: The general formulation of a quadratic equation is $ax^2+bx+c = 0$ with $a neq 0$. Let us say we are tasked with solving this equation, i.e., finding values of $x$ that satisfy this equation. To start with, note that this equation is easy to solve if $b=0$. Then the equation looks like $ax^2 + c = 0$ which would simplify to $x = pm sqrt{frac{-c}{a}}$.
"Completing the square" is a step that takes a general quadratic and reduces it to the form of the simple quadratic above. It does this with a substitution $y = x + frac{b}{2a}$. Then, $ay^2 = ax^2 + bx + frac{b^2}{4a}$ which means the general quadratic can be written as $ay^2-frac{b^2}{4a} + c = 0$ or equivalently as
$$ a(x+frac{b}{2a})^2 = frac{b^2}{4a} - c $$
This equation is easy to solve and yields the two roots of the general quadratic equation. Note that the above equation is equivalent to the one we started with ($ax^2+bx+c = 0$) for the purposes of finding the roots. This process of rewriting is called completing the square. This is the point behind rewriting $x^2+x+1$ as $(x+frac{1}{2})^2 + frac{3}{4}$.
answered Feb 5 '11 at 21:58
DineshDinesh
2,2391923
$endgroup$
"Completing the square" is a standard step in solving a quadratic equation. To see how it helps, consider the following: The general formulation of a quadratic equation is $ax^2+bx+c = 0$ with $a neq 0$. Let us say we are tasked with solving this equation, i.e., finding values of $x$ that satisfy this equation. To start with, note that this equation is easy to solve if $b=0$. Then the equation looks like $ax^2 + c = 0$ which would simplify to $x = pm sqrt{frac{-c}{a}}$.
"Completing the square" is a step that takes a general quadratic and reduces it to the form of the simple quadratic above. It does this with a substitution $y = x + frac{b}{2a}$. Then, $ay^2 = ax^2 + bx + frac{b^2}{4a}$ which means the general quadratic can be written as $ay^2-frac{b^2}{4a} + c = 0$ or equivalently as
$$ a(x+frac{b}{2a})^2 = frac{b^2}{4a} - c $$
This equation is easy to solve and yields the two roots of the general quadratic equation. Note that the above equation is equivalent to the one we started with ($ax^2+bx+c = 0$) for the purposes of finding the roots. This process of rewriting is called completing the square. This is the point behind rewriting $x^2+x+1$ as $(x+frac{1}{2})^2 + frac{3}{4}$.
answered Feb 5 '11 at 21:58
DineshDinesh
2,2391923
edited Feb 5 '11 at 22:22
answered Feb 5 '11 at 21:58
DineshDinesh
2,2391923
answered Feb 5 '11 at 21:58
DineshDinesh
2,2391923
answered Feb 5 '11 at 21:58
DineshDinesh
2,2391923
2,2391923
$begingroup$
"eliminating the middle term" is a bit misleading. One hasn't eliminated it. Rather one has "absorbed" it into the the square term by "completing" the "partial square" comprised of the two highest degree terms.
$endgroup$
– Bill Dubuque
Feb 5 '11 at 22:08
$begingroup$
@Bill: I agree. I will edit the answer accordingly. Thanks.
$endgroup$
– Dinesh
Feb 5 '11 at 22:16
$begingroup$
Arturo's answer explains the reason behind the name "completing the square" part very well. I will leave this answer up for the derivation of the roots of the quadratic part.
$endgroup$
– Dinesh
Feb 5 '11 at 22:22
add a comment |
$begingroup$
"eliminating the middle term" is a bit misleading. One hasn't eliminated it. Rather one has "absorbed" it into the the square term by "completing" the "partial square" comprised of the two highest degree terms.
$endgroup$
– Bill Dubuque
Feb 5 '11 at 22:08
$begingroup$
@Bill: I agree. I will edit the answer accordingly. Thanks.
$endgroup$
– Dinesh
Feb 5 '11 at 22:16
$begingroup$
Arturo's answer explains the reason behind the name "completing the square" part very well. I will leave this answer up for the derivation of the roots of the quadratic part.
$endgroup$
– Dinesh
Feb 5 '11 at 22:22
$begingroup$
"eliminating the middle term" is a bit misleading. One hasn't eliminated it. Rather one has "absorbed" it into the the square term by "completing" the "partial square" comprised of the two highest degree terms.
$endgroup$
– Bill Dubuque
Feb 5 '11 at 22:08
$begingroup$
"eliminating the middle term" is a bit misleading. One hasn't eliminated it. Rather one has "absorbed" it into the the square term by "completing" the "partial square" comprised of the two highest degree terms.
$endgroup$
– Bill Dubuque
Feb 5 '11 at 22:08
$begingroup$
@Bill: I agree. I will edit the answer accordingly. Thanks.
$endgroup$
– Dinesh
Feb 5 '11 at 22:16
$begingroup$
@Bill: I agree. I will edit the answer accordingly. Thanks.
$endgroup$
– Dinesh
Feb 5 '11 at 22:16
$begingroup$
Arturo's answer explains the reason behind the name "completing the square" part very well. I will leave this answer up for the derivation of the roots of the quadratic part.
$endgroup$
– Dinesh
Feb 5 '11 at 22:22
$begingroup$
Arturo's answer explains the reason behind the name "completing the square" part very well. I will leave this answer up for the derivation of the roots of the quadratic part.
$endgroup$
– Dinesh
Feb 5 '11 at 22:22
add a comment |
$begingroup$
One can rewrite a degree $rm:n > 1:$ polynomial $rm f(x) = x^n + b x^{n-1} + cdots $ into a form such that its two highest degree terms are "absorbed" into a perfect $rm: n$'th power of a linear polynomial, namely
$rmquadquad f(x) = (x + b/n)^n - g(x) $ where $rm g(x) = (x+b/n)^n - f(x) $ has degree $rm:le: n-2$
When $rm n = 2 $ this is called completing the square - esp. when used to solve a quadratic equation.$ $ If $rm g(x) = g $ is constant (as is always true when $rm n = 2:$) then this yields a closed form for the roots of $rm: f(x):, $ namely $rm x = sqrt[n]{g}-b/2:.$
answered Feb 5 '11 at 22:27
Bill DubuqueBill Dubuque
214k29197659
$endgroup$
add a comment |
$begingroup$
One can rewrite a degree $rm:n > 1:$ polynomial $rm f(x) = x^n + b x^{n-1} + cdots $ into a form such that its two highest degree terms are "absorbed" into a perfect $rm: n$'th power of a linear polynomial, namely
$rmquadquad f(x) = (x + b/n)^n - g(x) $ where $rm g(x) = (x+b/n)^n - f(x) $ has degree $rm:le: n-2$
When $rm n = 2 $ this is called completing the square - esp. when used to solve a quadratic equation.$ $ If $rm g(x) = g $ is constant (as is always true when $rm n = 2:$) then this yields a closed form for the roots of $rm: f(x):, $ namely $rm x = sqrt[n]{g}-b/2:.$
answered Feb 5 '11 at 22:27
Bill DubuqueBill Dubuque
214k29197659
$endgroup$
add a comment |
$begingroup$
One can rewrite a degree $rm:n > 1:$ polynomial $rm f(x) = x^n + b x^{n-1} + cdots $ into a form such that its two highest degree terms are "absorbed" into a perfect $rm: n$'th power of a linear polynomial, namely
$rmquadquad f(x) = (x + b/n)^n - g(x) $ where $rm g(x) = (x+b/n)^n - f(x) $ has degree $rm:le: n-2$
When $rm n = 2 $ this is called completing the square - esp. when used to solve a quadratic equation.$ $ If $rm g(x) = g $ is constant (as is always true when $rm n = 2:$) then this yields a closed form for the roots of $rm: f(x):, $ namely $rm x = sqrt[n]{g}-b/2:.$
answered Feb 5 '11 at 22:27
Bill DubuqueBill Dubuque
214k29197659
$endgroup$
One can rewrite a degree $rm:n > 1:$ polynomial $rm f(x) = x^n + b x^{n-1} + cdots $ into a form such that its two highest degree terms are "absorbed" into a perfect $rm: n$'th power of a linear polynomial, namely
$rmquadquad f(x) = (x + b/n)^n - g(x) $ where $rm g(x) = (x+b/n)^n - f(x) $ has degree $rm:le: n-2$
When $rm n = 2 $ this is called completing the square - esp. when used to solve a quadratic equation.$ $ If $rm g(x) = g $ is constant (as is always true when $rm n = 2:$) then this yields a closed form for the roots of $rm: f(x):, $ namely $rm x = sqrt[n]{g}-b/2:.$
answered Feb 5 '11 at 22:27
Bill DubuqueBill Dubuque
214k29197659
edited Feb 6 '11 at 0:46
user242
answered Feb 5 '11 at 22:27
Bill DubuqueBill Dubuque
214k29197659
answered Feb 5 '11 at 22:27
Bill DubuqueBill Dubuque
214k29197659
answered Feb 5 '11 at 22:27
Bill DubuqueBill Dubuque
214k29197659
214k29197659
add a comment |
add a comment |
$begingroup$
See this video and
its sequel to see the process worked in real time.
answered Feb 6 '11 at 16:06
Scott CarterScott Carter
2,2411113
$endgroup$
add a comment |
$begingroup$
See this video and
its sequel to see the process worked in real time.
answered Feb 6 '11 at 16:06
Scott CarterScott Carter
2,2411113
$endgroup$
add a comment |
$begingroup$
See this video and
its sequel to see the process worked in real time.
answered Feb 6 '11 at 16:06
Scott CarterScott Carter
2,2411113
$endgroup$
See this video and
its sequel to see the process worked in real time.
answered Feb 6 '11 at 16:06
Scott CarterScott Carter
2,2411113
answered Feb 6 '11 at 16:06
Scott CarterScott Carter
2,2411113
answered Feb 6 '11 at 16:06
Scott CarterScott Carter
2,2411113
answered Feb 6 '11 at 16:06
Scott CarterScott Carter
2,2411113
2,2411113
add a comment |
add a comment |
$begingroup$
As Arturo points out what you have to observe is the coefficient.
begin{align*}
x^{2}+x+1 & = x^{2} + x + frac{1+3}{4} \ & = x^{2} + x + frac{1}{4} + frac{3}{4} \ & = Bigl(x+frac{1}{2}Bigr)^{2} + frac{3}{4}
end{align*}
I am sure once you get used to such type of things you shall not have trouble in doing such problems. Solve more problems based on this type. Suppose you have the coefficient of $x$ as $a$ note that $a^{2}/4$ should be added and subtracted from the constant term. What i mean by this is: Suppose you have something of this type $x^{2}+ax + b^{2}$ then you can write this as $(x+a/2)^{2} + b^{2}-a^{2}/{4}$.
$endgroup$
add a comment |
$begingroup$
As Arturo points out what you have to observe is the coefficient.
begin{align*}
x^{2}+x+1 & = x^{2} + x + frac{1+3}{4} \ & = x^{2} + x + frac{1}{4} + frac{3}{4} \ & = Bigl(x+frac{1}{2}Bigr)^{2} + frac{3}{4}
end{align*}
I am sure once you get used to such type of things you shall not have trouble in doing such problems. Solve more problems based on this type. Suppose you have the coefficient of $x$ as $a$ note that $a^{2}/4$ should be added and subtracted from the constant term. What i mean by this is: Suppose you have something of this type $x^{2}+ax + b^{2}$ then you can write this as $(x+a/2)^{2} + b^{2}-a^{2}/{4}$.
$endgroup$
add a comment |
$begingroup$
As Arturo points out what you have to observe is the coefficient.
begin{align*}
x^{2}+x+1 & = x^{2} + x + frac{1+3}{4} \ & = x^{2} + x + frac{1}{4} + frac{3}{4} \ & = Bigl(x+frac{1}{2}Bigr)^{2} + frac{3}{4}
end{align*}
I am sure once you get used to such type of things you shall not have trouble in doing such problems. Solve more problems based on this type. Suppose you have the coefficient of $x$ as $a$ note that $a^{2}/4$ should be added and subtracted from the constant term. What i mean by this is: Suppose you have something of this type $x^{2}+ax + b^{2}$ then you can write this as $(x+a/2)^{2} + b^{2}-a^{2}/{4}$.
$endgroup$
As Arturo points out what you have to observe is the coefficient.
begin{align*}
x^{2}+x+1 & = x^{2} + x + frac{1+3}{4} \ & = x^{2} + x + frac{1}{4} + frac{3}{4} \ & = Bigl(x+frac{1}{2}Bigr)^{2} + frac{3}{4}
end{align*}
I am sure once you get used to such type of things you shall not have trouble in doing such problems. Solve more problems based on this type. Suppose you have the coefficient of $x$ as $a$ note that $a^{2}/4$ should be added and subtracted from the constant term. What i mean by this is: Suppose you have something of this type $x^{2}+ax + b^{2}$ then you can write this as $(x+a/2)^{2} + b^{2}-a^{2}/{4}$.
answered Feb 5 '11 at 22:14
anonymous
add a comment |
add a comment |
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$begingroup$
en.wikipedia.org/wiki/Completing_the_square
$endgroup$
– Dinesh
Feb 5 '11 at 21:48