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sologuitar

Member

Registered: 2022-09-19

Posts: 467

Completing the Square

Solve equation by completing the Square.

2x^2 + π x + 2 = 0

Let me see.

In order to complete the square, the coefficient of the x square term
cannot be greater than or less than 1.

So, I must divide each term on both sides of the equation, in this case, by 2.

(2x^2 + π x + 2)/2 = 0/2

Doing so, I get the following:

x^2 + (π/2) x + 1 = 0

I now subtract 1 from both sides.

x^2 + (π/2) x = -1

The next step is to raise the coefficient of x to the second power and add the product to both sides of the equation.

(π/2)^2 = π^2/4

x^2 + (π/2) x + π^2/4  = -1 + π^2/4

Stuck here....

Bob

Administrator

Registered: 2010-06-20

Posts: 10,190

Re: Completing the Square

hi

That's a bit confused.

You're right to make the x^2 coefficient 1. Then you need to add something to make the correct constant term.

Take a look at a perfect square:

So if your quadratic looks like this:

.....(1)

you need to add a^2

Here's an example:

Comparing with (1) I can see that a=2

a^2 = 4 so that's what I need to add (12 to each side)

The left hand side is a perfect square so I can make this factorisation

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

sologuitar

Member

Registered: 2022-09-19

Posts: 467

Re: Completing the Square

hi

That's a bit confused.

You're right to make the x^2 coefficient 1. Then you need to add something to make the correct constant term.

Take a look at a perfect square:

So if your quadratic looks like this:

.....(1)

you need to add a^2

Here's an example:

Comparing with (1) I can see that a=2

a^2 = 4 so that's what I need to add (12 to each side)

The left hand side is a perfect square so I can make this factorisation

Bob

I added to both sides in my question. Do you see it?

Look here:

Given (π/2)^2 = π^2/4, I now must add π^2/4 to both sides of the equation.

Here it is:

x^2 + (π/2) x + π^2/4  = -1 + π^2/4

Where do I go from here?

amnkb

Member

Registered: 2023-09-19

Posts: 253

Re: Completing the Square

Given (π/2)^2 = π^2/4, I now must add π^2/4 to both sides of the equation.

Here it is:

x^2 + (π/2) x + π^2/4  = -1 + π^2/4

Where do I go from here?

turn the left-hand side into perfect-square form
combine the right-hand side to being 1 fraction
take the square root of both sides
solve for x= on the left-hand side

sologuitar

Member

Registered: 2022-09-19

Posts: 467

Re: Completing the Square

Given (π/2)^2 = π^2/4, I now must add π^2/4 to both sides of the equation.

Here it is:

x^2 + (π/2) x + π^2/4  = -1 + π^2/4

Where do I go from here?

turn the left-hand side into perfect-square form
combine the right-hand side to being 1 fraction
take the square root of both sides
solve for x= on the left-hand side

x^2 + (π/2) x + π^2/4  = -1 + π^2/4

(x + (π/2))(x + (π/2)) = 1.4674

(x + (π/2))^2 = 1.4674

I now take the square root on both sides. Yes?

amnkb

Member

Registered: 2023-09-19

Posts: 253

Re: Completing the Square

x^2 + (π/2) x + π^2/4  = -1 + π^2/4

(x + (π/2))(x + (π/2)) = 1.4674

(x + (π/2))^2 = 1.4674

I now take the square root on both sides. Yes?

where did 1.4674 come from?
did you assign n the value of pi?
how did you get the square of n/4 being (n^2)/4?
shouldnt the 4 be squared too?
not understanding your results
sry

sologuitar

Member

Registered: 2022-09-19

Posts: 467

Re: Completing the Square

x^2 + (π/2) x + π^2/4  = -1 + π^2/4

(x + (π/2))(x + (π/2)) = 1.4674

(x + (π/2))^2 = 1.4674

I now take the square root on both sides. Yes?

where did 1.4674 come from?
did you assign n the value of pi?
how did you get the square of n/4 being (n^2)/4?
shouldnt the 4 be squared too?
not understanding your results
sry

Look:

-1 + π^2/4 = 1.4674 according to the calculator.

amnkb

Member

Registered: 2023-09-19

Posts: 253

Re: Completing the Square

where did 1.4674 come from?
did you assign n the value of pi?
how did you get the square of n/4 being (n^2)/4?
shouldnt the 4 be squared too?
not understanding your results
sry

Look:

-1 + π^2/4 = 1.4674 according to the calculator.

so n equals

?

also the middle coefficient of your equation is (pi)/2
you're supposed to take half of this and then square it to get the third term to complete the square
the complete-the-squarr process is:

half of b/a is b/2a
square this to get the completed-square 3erd term:

half of (pi)/2 is (pi)/4
square this to get ((pi)^2)/16
unless they tell you specifically to round to 4 decimal places you should do exact using pi

sologuitar

Member

Registered: 2022-09-19

Posts: 467

Re: Completing the Square

where did 1.4674 come from?
did you assign n the value of pi?
how did you get the square of n/4 being (n^2)/4?
shouldnt the 4 be squared too?
not understanding your results
sry

Look:

-1 + π^2/4 = 1.4674 according to the calculator.

so n equals

?

also the middle coefficient of your equation is (pi)/2
you're supposed to take half of this and then square it to get the third term to complete the square
the complete-the-squarr process is:

half of b/a is b/2a
square this to get the completed-square 3erd term:

half of (pi)/2 is (pi)/4
square this to get ((pi)^2)/16
unless they tell you specifically to round to 4 decimal places you should do exact using pi

Ok. I see that my first mistake was squaring wrongly.
Can you finish this one for me?

sologuitar

Member

Registered: 2022-09-19

Posts: 467

Re: Completing the Square

where did 1.4674 come from?
did you assign n the value of pi?
how did you get the square of n/4 being (n^2)/4?
shouldnt the 4 be squared too?
not understanding your results
sry

Look:

-1 + π^2/4 = 1.4674 according to the calculator.

so n equals

?

also the middle coefficient of your equation is (pi)/2
you're supposed to take half of this and then square it to get the third term to complete the square
the complete-the-squarr process is:

half of b/a is b/2a
square this to get the completed-square 3erd term:

half of (pi)/2 is (pi)/4
square this to get ((pi)^2)/16
unless they tell you specifically to round to 4 decimal places you should do exact using pi

x^2 + (π/2) x + π^2/16  = -1 + π^2/16

(x + (π/4))(x + (π/4)) = 1.4674

(x + (π/4))^2 = -0.38315

Stuck here....

amnkb

Member

Registered: 2023-09-19

Posts: 253

Re: Completing the Square

the complete-the-squarr process is:

half of b/a is b/2a
square this to get the completed-square 3erd term
add this squared term to both sides
including any multiplier 'a':

half of (pi)/2 is (pi)/4
square this to get ((pi)^2)/16
unless they tell you specifically to round to 4 decimal places you should do exact using pi

x^2 + (π/2) x + π^2/16  = -1 + π^2/16

(x + (π/4))(x + (π/4)) = 1.4674

like bob said: stop rounding
use exact values

(x + (π/4))^2 = -0.38315

where did -0.38315 come from?
all you did was change (expression)(expression) to (expression)^2
thats just a change of formatting
the value of the other side should not have changed

Stuck here....

follow the steps exactly as listed
including adding the (half of (middle coefficient))^2 to both sides

sologuitar

Member

Registered: 2022-09-19

Posts: 467

Re: Completing the Square

the complete-the-squarr process is:

half of b/a is b/2a
square this to get the completed-square 3erd term
add this squared term to both sides
including any multiplier 'a':

half of (pi)/2 is (pi)/4
square this to get ((pi)^2)/16
unless they tell you specifically to round to 4 decimal places you should do exact using pi

x^2 + (π/2) x + π^2/16  = -1 + π^2/16

(x + (π/4))(x + (π/4)) = 1.4674

like bob said: stop rounding
use exact values

(x + (π/4))^2 = -0.38315

where did -0.38315 come from?
all you did was change (expression)(expression) to (expression)^2
thats just a change of formatting
the value of the other side should not have changed

Stuck here....

follow the steps exactly as listed
including adding the (half of (middle coefficient))^2 to both sides

Ok. I will work it out completely later and show my work here.

sologuitar

Member

Registered: 2022-09-19

Posts: 467

Re: Completing the Square

the complete-the-squarr process is:

half of b/a is b/2a
square this to get the completed-square 3erd term
add this squared term to both sides
including any multiplier 'a':

half of (pi)/2 is (pi)/4
square this to get ((pi)^2)/16
unless they tell you specifically to round to 4 decimal places you should do exact using pi

x^2 + (π/2) x + π^2/16  = -1 + π^2/16

(x + (π/4))(x + (π/4)) = 1.4674

like bob said: stop rounding
use exact values

(x + (π/4))^2 = -0.38315

where did -0.38315 come from?
all you did was change (expression)(expression) to (expression)^2
thats just a change of formatting
the value of the other side should not have changed

Stuck here....

follow the steps exactly as listed
including adding the (half of (middle coefficient))^2 to both sides

Solve equation by completing the Square.

2x^2 + π x + 2 = 0

Let me see.

In order to complete the square, the coefficient of the x square term
cannot be greater than or less than 1.

So, I must divide each term on both sides of the equation, in this case, by 2.

(2x^2 + π x + 2)/2 = 0/2

Doing so, I get the following:

x^2 + (π/2) x + 1 = 0

I now subtract 1 from both sides.

x^2 + (π/2) x = -1

The next step is to divide the coefficient of x by 2 and then raise to the second power and add the product to both sides of the equation.

[(π/2)^2]/2 = (π/4)^2 = (π^2/16)

I must now add the product to both sides of the equation.

x^2 + (π/2) x + π^2/16  = -1 + π^2/16

x^2 + (π/2) x + π/4 = -1 + π^2/16

(x + π/4)(x + π/4) = negative answer.  Do you understand why I needed to subtract 1 from both sides of the equation a few steps before? Do you know where -1 comes from? I showed where -1 comes from a few steps before. Now, with the right side being negative, can we conclude that this equation has no solution over the real numbers?

You say?

Bob

Administrator

Registered: 2010-06-20

Posts: 10,190

Re: Completing the Square

It does look that way. If you put those numbers into this:

https://www.mathsisfun.com/quadratic-eq … olver.html

The graph doesn't cross the x axis which would confirm no real roots.

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

sologuitar

Member

Registered: 2022-09-19

Posts: 467

Re: Completing the Square

It does look that way. If you put those numbers into this:

https://www.mathsisfun.com/quadratic-eq … olver.html

The graph doesn't cross the x axis which would confirm no real roots.

Bob

Ok. No solution over the real numbers. Cool. My concern is to complete the square correctly. Thanks again.

Note:

Bob,

My family and friends are confused about love for mathematics. They say that solving textbook questions at 58 is a clear indication that something may be wrong with me. Really? What can I do at 58? I also play classical guitar hymns. They are also confused about my love for music and playing guitar.

At 58:

A. I am not popular with women anymore

B. Bald

C. Terribly overweight

D. Work For a dead-end security company

Do you ever come across people who have nothing else to do but criticize something you love to do? If math brings me joy, what's wrong with that? If playing guitar brings me joy, what's wrong with that?

You say?