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This module teaches the method of solving quadratic equations by completing the square.

Consider the equation:

x + 3 2 = 16 size 12{ left (x+3 right ) rSup { size 8{2} } ="16"} {}

We can solve this by analogy to the way that we approached absolute value problems. something squared is 16. So what could the something be? It could be 4. It could also be 4 size 12{ - 4} {} . So the solution is:

  • x + 3 = 4 size 12{x+3=4} {}
    • x = 1 size 12{x=1} {}
  • x + 3 = 4 size 12{x+3= - 4} {}
    • x = 7 size 12{x= - 7} {}

These are the two solutions.

This simple problem leads to a completely general way of solving quadratic equations—because any quadratic equation can be put in a form like the above equation. The key is completing the square which, in turn, is based on our original two formulae:

x + a 2 = x 2 + 2 ax + a 2 size 12{ left (x+a right ) rSup { size 8{2} } =x rSup { size 8{2} } +2 ital "ax"+a rSup { size 8{2} } } {}
x a 2 = x 2 2 ax + a 2 size 12{ left (x - a right ) rSup { size 8{2} } =x rSup { size 8{2} } - 2 ital "ax"+a rSup { size 8{2} } } {}

As an example, consider the equation x 2 + 10 x + 21 = 0 size 12{x rSup { size 8{2} } +"10"x+"21"=0} {} . In order to make it fit one of the patterns above, we must replace the 21 with the correct number: a number such that x 2 + 10 x + __ size 12{x rSup { size 8{2} } +"10"x+"__"} {} is a perfect square. What number goes there? If you are familiar with the pattern, you know the answer right away. 10 is 5 doubled , so the number there must be 5 squared , or 25.

But how do we turn a 21 into a 25? We add 4, of course. And if we add 4 to one side of the equation, we have to add 4 to the other side of the equation. So the entire problem is worked out as follows:

Solving by Completing the Square (quick-and-dirty version)
Solve x 2 + 10 x + 21 = 0 size 12{x rSup { size 8{2} } +"10"x+"21"=0} {} The problem.
x 2 + 10 x + 25 = 4 size 12{x rSup { size 8{2} } +"10"x+"25"=4} {} Add 4 to both sides, so that the left side becomes a perfect square.
x + 5 2 = 4 size 12{ left (x+5 right ) rSup { size 8{2} } =4} {} Rewrite the perfect square.
x + 5 = 2 x = 3 size 12{ matrix { x+5=2 {} ##x= - 3 } } {} x + 5 = 2 x = 7 size 12{ matrix { x+5= - 2 {} ##x= - 7 } } {} If something-squared is 4, the something can be 2, or 2 size 12{ - 2} {} . Solve both possibilities to find the two answers.

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Thus, we have our two solutions.

Completing the square is more time-consuming than factoring: so whenever a quadratic equation can be factored, factoring is the preferred method. (In this case, we would have factored the original equation as x + 3 x + 7 size 12{ left (x+3 right ) left (x+7 right )} {} and gotten straight to the answer.) However, completing the square can be used on any quadratic equation. In the example below, completing the square is used to find two answers that would not have been found by factoring.

Solving by Completing the Square (showing all the steps more carefully)
Solve 9x 2 54 x + 80 = 0 size 12{9x rSup { size 8{2} } - "54"x+"80"=0} {} The problem.
9x 2 54 x = 80 size 12{9x rSup { size 8{2} } - "54"x= - "80"} {} Put all the x size 12{x} {} terms on one side, and the number on the other
x 2 6x = 80 9 size 12{x rSup { size 8{2} } - 6x= - { {"80"} over {9} } } {} Divide both sides by the coefficient of x 2 size 12{x rSup { size 8{2} } } {} (*see below)
x 2 6x + 9 ̲ = 80 9 + 9 ̲ size 12{x rSup { size 8{2} } - 6x {underline {+9}} = - { {"80"} over {9} } + {underline {9}} } {} Add the same number (*see below) to both sides, so that the left side becomes a perfect square.
x 3 2 = 80 9 + 81 9 = 1 9 size 12{ left (x - 3 right ) rSup { size 8{2} } = - { {"80"} over {9} } + { {"81"} over {9} } = { {1} over {9} } } {} Rewrite the perfect square.
x 3 = 1 3 x = 3 1 3 size 12{ matrix { x - 3= { {1} over {3} } {} ##x=3 { { size 8{1} } over { size 8{3} } } } } {} x 3 = 1 3 x = 2 2 3 size 12{ matrix { x - 3= - { {1} over {3} } {} ##x=2 { { size 8{2} } over { size 8{3} } } } } {} If something-squared is 1 9 size 12{ { {1} over {9} } } {} , the something can be 1 3 size 12{ { {1} over {3} } } {} , or 1 3 size 12{ - { {1} over {3} } } {} . Solve both possibilities to find the two answers.

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Two steps in particular should be pointed out here.

In the third step, we divide both sides by 9. When completing the square, you do not want to have any coefficient in front of the term; if there is a number there, you divide it out. Fractions, on the other hand (such as the 80 9 size 12{ - { {"80"} over {9} } } {} in this case) do not present a problem. This is in marked contrast to factoring, where a coefficient in front of the x 2 size 12{x rSup { size 8{2} } } {} can be left alone, but fractions make things nearly impossible.

The step after that is where we actually complete the square. x 2 + 6x + __ size 12{x rSup { size 8{2} } +6x+"__"} {} will be our perfect square. How do we find what number we want? Start with the coefficient of x 2 size 12{x rSup { size 8{2} } } {} (in this case, 6). Take half of it, and square the result . Half of 6 is 3, squared is 9. So we want a 9 there to create x 2 + 6x + 9 size 12{x rSup { size 8{2} } +6x+9} {} which can be simplified to x + 3 2 size 12{ left (x+3 right ) rSup { size 8{2} } } {} .

An image showing how to select the correct number and position it within the equation correctly when completing the square.

If the coefficient of x size 12{x} {} is an odd number , the problem becomes a little uglier, but the principle is the same. For instance, faced with:

x 2 + 5x + __ size 12{x rSup { size 8{2} } +5x+"__"} {}

You would begin by taking half of 5 (which is 5 2 size 12{ { {5} over {2} } } {} ) and then squaring it:

x 2 + 5x + 25 4 = x + 5 2 2 size 12{x rSup { size 8{2} } +5x+ { {"25"} over {4} } = left (x+ { {5} over {2} } right ) rSup { size 8{2} } } {}

Another “completing the square” example, in which you cannot get rid of the square root at all, is presented in the worksheet “The Generic Quadratic Equation.”

One final note on completing the square: there are three different possible outcomes.

  • If you end up with something like x 3 2 = 16 size 12{ left (x - 3 right ) rSup { size 8{2} } ="16"} {} you will find two solutions, since x 3 size 12{x - 3} {} can be either 4, or 4 size 12{ - 4} {} . You will always have two solutions if the right side of the equation, after completing the square, is positive.
  • If you end up with x 3 2 = 0 size 12{ left (x - 3 right ) rSup { size 8{2} } =0} {} then there is only one solution: x must be 3 in this example. If the right side of the equation is 0 after completing the square, there is only one solution.
  • Finally, what about a negative number on the right, such as x 3 2 = 16 size 12{ left (x - 3 right ) rSup { size 8{2} } = - "16"} {} ? Nothing squared can give a negative answer, so there is no solution.

Questions & Answers

find the 15th term of the geometric sequince whose first is 18 and last term of 387
Jerwin Reply
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
virgelyn Reply
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Abhi
how do they get the third part x = (32)5/4
kinnecy Reply
can someone help me with some logarithmic and exponential equations.
Jeffrey Reply
sure. what is your question?
ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
hmm
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salma
Commplementary angles
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what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
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or infinite solutions?
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The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
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I start with an easy one. carbon nanotubes woven into a long filament like a string
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Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
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Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
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silver nanoparticles could handle the job?
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I'm interested in Nanotube
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Prasenjit Reply
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
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the Beer law works very well for dilute solutions but fails for very high concentrations. why?
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how did you get the value of 2000N.What calculations are needed to arrive at it
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Source:  OpenStax, Advanced algebra ii: conceptual explanations. OpenStax CNX. May 04, 2010 Download for free at http://cnx.org/content/col10624/1.15
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