Basic Completing the Square Question

[askor]

How do you complete the square of this?:

$x^2 - 3x - 7$

I already worked on this but the result is not integer:

$(x - 1.5)^2 - (\sqrt{9.25})^2$

Is my above work correct?

How do you obtain the result in integer?

How do you complete the square so that the result is integer?

For example:
$x^2 - x - any constant$
$x^2 - 3x - any constant$
$x^2 - 5x - any constant$
$x^2 - 7x - any constant$
$x^2 - 9x - any constant$

and so on, where the magnitude of $x$ is odd.

 

[fresh_42]

It is correct, and there is no guarantee for integers. If you apply the method on $x^2+px+q=0$ then you get none, one, or two real solutions. They are usually not integer solutions.

If you are happy and you can guess a solution, say $a$ which satisfies $a^2+pa+q=0$ then you can divide $(x^2+px+q):(x-a)$ by long division and get the other solution (or none, if there is no real solution).

 

[askor]

...there is no guarantee for integers.

What do you mean?

 

[fresh_42]

What do you mean?

$x^2-3x-7=\left(x-\dfrac{3}{2}-\dfrac{\sqrt{37}}{2}\right)\cdot \left(x-\dfrac{3}{2}+\dfrac{\sqrt{37}}{2}\right)$ so there is no way to get rid of the quotient, nor the square root.

 

[askor]

$x^2-3x-7=\left(x-\dfrac{3}{2}-\dfrac{\sqrt{37}}{2}\right)\cdot \left(x-\dfrac{3}{2}+\dfrac{\sqrt{37}}{2}\right)$ so there is no way to get rid of the quotient, nor the square root.

Do you mean $\frac{\sqrt{7}}{2}$?

 

[DaveE]

What do you mean?

Let's look at this in the reverse direction. Pick any number, let's call it n, it doesn't have to be an integer. You can make a square polynomial (x-n)² = x²-2nx+n² with it.

Now any polynomial with the form x²-2nx+k (for any other constant k) can be expressed as (x-n)²+(k-n²). This is completing the square. There is no need to limit the type of constants allowed.

For example x²+6πx+3/4 = (x+3π)²+3/4-9π², with n=-3π, k=3/4.

 

[fresh_42]

No. I meant what I wrote. We have $x^2-3x-7$ and want to compare it with $(x-a)^2+b=x^2-2ax+a^2+b.$ So - as you correctly did - we get
$$
x^2\quad\underbrace{-3}_{=-2a}x\quad\underbrace{-7}_{=a^2+b} \Longrightarrow a=-1.5\text{ and }-7=(1.5)^2+b\text{ or }b=-7-(1.5)^2=-9.25
$$
so $x^2-3x-7=(x-1.5)^2-9.25$ which you had, too.

 

[symbolipoint]

Putting the numeric example into a geometric rectangle & square approach will make the process of Completing The Square, very very clear. If you cannot find this or cannot figure it on your own, say, and someone will provide something.

 

[askor]

Putting the numeric example into a geometric rectangle & square approach will make the process of Completing The Square, very very clear. If you cannot find this or cannot figure it on your own, say, and someone will provide something.

Yes, please show me, I am interested.

 

[DaveE]

Yes, please show me, I am interested.

https://www.mathsisfun.com/algebra/completing-square.html

 

[Charles Link]

To complete the square of $x^2 +bx+c$, you add and subtract $(b/2)^2$. The important thing is to get $x^2+bx+c=(x+\frac{b}{2})^2+c-(\frac{b}{2})^2$. You usually don't need to put it in the form $(x+e)^2-d^2$.

The $-d^2$ term is often better left simply as $c-(\frac{b}{2})^2$.

The $x^2+bx$ that is at the core of this operation. To make it a square, you add $(\frac{b}{2})^2$.

 

[epenguin]

You know $\left( x+a\right) ^{2}=x^{2}+2ax+a^{2}$ ?
So to get the $x$ term inside a square of ($x$+ something) your 3 will have to equal $2a$, so $a=\dfrac {3}{2}$ and the expression is $\left( x+\dfrac {3}{2}\right) ^{2}-\dfrac {1}{4}$ or if preferred
$$\left( x+\dfrac {3}{2}\right) ^{2}-\left( \dfrac {1}{2}\right) ^{2}$$
I wonder at what school you never saw anything like that?
I know you got it in the end, seen in an exercise on indefinite integration, where you have to get the integrand in the form
$$\dfrac {1}{\sqrt {\left( u^{2}-1\right) }}$$
That and similar problems by the way are in every calculus textbook with a chapter on indefinite integration - one of the half dozen or so standard forms which with their variations and disguises make up the standard forms frequently encountered and that can actually be integrated in terms of standard functions.

 

[symbolipoint]

Yes, please show me, I am interested.

Please try a search on the internet, or especially a search on YouTube.

Here is one which looks ok based on just the thumbnail:


The part YOU want to see begins at about 1:50 on the timeline.

 

[PeroK]

How do you complete the square of this?:

$x^2 - 3x - 7$

I already worked on this but the result is not integer:

$(x - 1.5)^2 - (\sqrt{9.25})^2$

Is my above work correct?

How do you obtain the result in integer?

Here's a question. We have a quadratic equation $$ax^2 + bx + c$$ with up to two solutions given by $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ Does that quadratic formula apply (assuming $a \ne 0$):

1) For any integers $a, b, c$.

2) For any rational numbers $a, b, c$.

3) For any real numbers $a, b, c$