How Do You Complete the Square for x^2 + 3x + 4?

In summary, the conversation is about completing the square for an equation with fractions. The method of completing the square is discussed and it is noted that decimals can also be used. The process is explained and it is mentioned that fractions can be confusing for some people.
  • #1
iNsChris
22
0
Complete the square - Help please :)

Not to good with fractions, so I'm posting this here.


x^2 + 3x +4

Complete the square.

Ill try:

(x + 3/6)^2 + ?

lol best i can do sorry :)

Thanks
 
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  • #2
iNsChris said:
Not to good with fractions, so I'm posting this here.


x^2 + 3x +4

Complete the square.

Ill try:

(x + 3/6)^2 + ?

lol best i can do sorry :)

Thanks

x^2 + 3x + 4
=> (x + 1.5)^2 +4 - 2.25
=> (x + 1.5)^2 +1.75

There you go.

The Bob (2004 ©)
 
  • #3
yeh i considered decimals but book showed fractions - Cheers mate.
 
  • #4
iNsChris said:
yeh i considered decimals but book showed fractions - Cheers mate.

What is that supposed to mean? It makes no difference how you express the numbers (every fraction has some decimal representation!), the method of solving the problem is always the same.

[tex] x^2 + 3x +4 = \left(x + \frac{3}{2}\right)^2 + 4 - \left(\frac{3}{2}\right)^2 [/tex]

Do you see the general procedure for completing the square? It's always the same. Think about what it means to "complete the square". We want to add 'something' to the expression so that we end up with a perfect square. (and then subtract that same 'something' so that in the end we haven't changed anything). i.e. so that

x^2 + 3x + something + 4 - something

can be expressed using the square of some binomial (ie so that the italicized part is a perfect square):

= (x + ?)^2 +4 - something

Can you see that '?' must be half of 3 in this case? -- because when you expand, you get x^2 + 2?x + ?^2
so (2? = 3)

It follows that 'something' = ?^2 = (3/2)^2 = 9/4

So if you understand and follow this reasoning every single time, completing the square will never be any trouble.


btw last time I checked:

3/2 = 1.5

9/4 = 2.25

:wink:
 
  • #5
for the function, actually there's an easier way to complete the square, just simply expand it out, what i mean is

[tex]x^3+3x+4=a(x+b)^2+c[/tex]

equate these 2 together, expand out [tex]a(x+b)^2+c[/tex] then compare the coefficients.
 
  • #6
I believe Chris understands this but thought that he had to use fractions when really decimals were alright.

The completing the square process is simple and he has it but it was a little doubt that needed sorting.

The Bob (2004 ©)
 
  • #7
Bob's right. Fractions always get people. It took me quite a while to get used to them myself.
 

FAQ: How Do You Complete the Square for x^2 + 3x + 4?

What is the purpose of completing the square?

Completing the square is a method used in algebra to solve quadratic equations. It allows us to manipulate the equation into a perfect square form, making it easier to solve for the variable.

How do I complete the square?

To complete the square, follow these steps:

  1. Move the constant term to the right side of the equation.
  2. Divide the coefficient of the x-term by 2 and square it.
  3. Add this value to both sides of the equation.
  4. Factor the perfect square on the left side, and simplify the right side.
  5. Take the square root of both sides to solve for the variable.

When should I use completing the square?

Completing the square is typically used to solve quadratic equations that cannot be factored easily. It can also be used to find the vertex of a parabola or to graph a quadratic equation.

What are the benefits of completing the square?

Completing the square allows us to solve for a variable in a quadratic equation, even if the equation cannot be factored. It also helps us find the vertex of a parabola, which is useful in graphing quadratic equations.

Are there any limitations when using completing the square?

Completing the square can only be used to solve quadratic equations. It also involves some algebraic manipulation, which can be challenging for some students. Additionally, it may not always be the most efficient method for solving quadratic equations.

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