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Quadratic Equations

Casio

Member
Feb 11, 2012
86
I have been reading up on Quadratic Equations in my course book, which gives some examples of simple equations and how they are factorised, but can't get my head round this type which is has no examples.

x2 - 3x = 0

OK I know it is a quadratic because the term x2 is included. What I require is a value for the term x.

Is there a method to factorise this I am not sure?

x2 - 3x = x(x - 3)

It does not give me a value for x?

I can assume values like x = 0 and x = 3, both of which will = 0, but that does not show me how I get the method of factorising this equation?

Thanks

Casio
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,193
There is a way of factoring quadratics that takes the guesswork out of it. In your example, you just note that $x$ is common to both terms, and factors out leaving you with $x(x-3)=0$.

But let's suppose you had a general quadratic: $x^{2}+bx+c$, which we'll make monic for convenience. What you're after is an equivalent expression that looks like this:
$$x^{2}+bx+c=(x+f)(x+g).$$
So, FOIL out the RHS thus:
$$x^{2}+bx+c=x^{2}+gx+fx+fg=x^{2}+(g+f)x+fg.$$
Now, you get to equate like powers of $x$ to obtain the following equations:
\begin{align*}
b&=f+g\\
c&=fg.
\end{align*}
Use whatever method you like for solving this simultaneous system of equations for $f$ and $g$. Once you do that, you're done.

Example: Factor $x^{2}+8x-3$.

Solution: Write as
$$x^{2}+8x-3=(x+f)(x+g)=x^{2}+(f+g)x+fg.$$
Solve:
\begin{align*}
8&=f+g\\
-3&=fg.
\end{align*}
The solution is
$$f=4\pm\sqrt{19},\quad g=4\mp\sqrt{19}.$$
That is, if $f=4+\sqrt{19}$, then $g=4-\sqrt{19}$, or if $f=4-\sqrt{19}$, then $g=4+\sqrt{19}$. So our factorization is
$$x^{2}+8x-3=(x+4-\sqrt{19})(x+4+\sqrt{19}).$$
Multiply this out to confirm.

Does that help?
 
Jul 22, 2012
35

SuperSonic4

Well-known member
MHB Math Helper
Mar 1, 2012
249
How did you solve this?
Simultaneous equations: there are two variables and two equations. In the terms of a quadratic you may know it along the lines of "two numbers which add to 8 and multiply to -3"

If you need more help on them feel free to create a new topic, it's not fair on Casio to sidetrack his topic :)
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,193
How did you solve this?
Substitution. We have that $f=8-g$, so plugging that into the other equation yields
$$-3=g(8-g)=8g-g^{2},$$
or
$$g^{2}-8g-3=0.$$
Now you have a quadratic in $g$. You can use the quadratic formula to solve:
$$g=\frac{8\pm\sqrt{64-4(-3)}}{2}=\dots$$
Come to think of it, we haven't gained anything here, because this is the same quadratic as we started with!

So you could save time by simply using the quadratic formula in the first place. Given the monic quadratic $x^{2}+bx+c$, factor by finding the roots:
$$x=\frac{-b\pm\sqrt{b^{2}-4c}}{2}.$$
Factoring the original quadratic would then give you
$$x^{2}+bx+c=\left(x+\frac{b+\sqrt{b^{2}-4c}}{2}\right)\left(x+\frac{b-\sqrt{b^{2}-4c}}{2}\right).$$
 
Jul 22, 2012
35
If you need more help on them feel free to create a new topic, it's not fair on Casio to sidetrack his topic :)
No need to be rude here. The question I have asked is very relevant. These simultaneous equations are the Vieta relations of the quadratic equation being solved. Trying to eliminate one variable would get you back to the original equation.
 
Jul 22, 2012
35
Substitution. We have that $f=8-g$, so plugging that into the other equation yields
$$-3=g(8-g)=8g-g^{2},$$
or
$$g^{2}-8g-3=0.$$
Now you have a quadratic in $g$. You can use the quadratic formula to solve:
$$g=\frac{8\pm\sqrt{64-4(-3)}}{2}=\dots$$
Come to think of it, we haven't gained anything here, because this is the same quadratic as we started with!

So you could save time by simply using the quadratic formula in the first place. Given the monic quadratic $x^{2}+bx+c$, factor by finding the roots:
$$x=\frac{-b\pm\sqrt{b^{2}-4c}}{2}.$$
Factoring the original quadratic would then give you
$$x^{2}+bx+c=\left(x+\frac{b+\sqrt{b^{2}-4c}}{2}\right)\left(x+\frac{b-\sqrt{b^{2}-4c}}{2}\right).$$
I was thinking about this, and I thought what if we try to find $f-g$ first?

$(f-g)^2 = (f+g)^2-4fg \implies f-g = \pm \sqrt{76}$.


$\left\{\begin{array}{c}f+g = 8 \\\\ f-g = \pm\sqrt{76} \end{array} \implies 2f = 8\pm \sqrt{76} \implies \boxed{f = 4\pm\frac{1}{2}\sqrt{76}} \implies \boxed{g = 4\mp\frac{1}{2}\sqrt{76}}.\right\|$

Without loss of generality, we take $f = 4+\frac{1}{2}\sqrt{76}$ and $g = 4-\frac{1}{2}\sqrt{76}$.

I think this also gives a way of proving the quadratic formula!
 
Last edited:

Casio

Member
Feb 11, 2012
86
There is a way of factoring quadratics that takes the guesswork out of it. In your example, you just note that $x$ is common to both terms, and factors out leaving you with $x(x-3)=0$.

But let's suppose you had a general quadratic: $x^{2}+bx+c$, which we'll make monic for convenience. What you're after is an equivalent expression that looks like this:
$$x^{2}+bx+c=(x+f)(x+g).$$
So, FOIL out the RHS thus:
$$x^{2}+bx+c=x^{2}+gx+fx+fg=x^{2}+(g+f)x+fg.$$
Now, you get to equate like powers of $x$ to obtain the following equations:
\begin{align*}
b&=f+g\\
c&=fg.
\end{align*}
Use whatever method you like for solving this simultaneous system of equations for $f$ and $g$. Once you do that, you're done.

Example: Factor $x^{2}+8x-3$.

Solution: Write as
$$x^{2}+8x-3=(x+f)(x+g)=x^{2}+(f+g)x+fg.$$
Solve:
\begin{align*}
8&=f+g\\
-3&=fg.
\end{align*}
The solution is
$$f=4\pm\sqrt{19},\quad g=4\mp\sqrt{19}.$$
That is, if $f=4+\sqrt{19}$, then $g=4-\sqrt{19}$, or if $f=4-\sqrt{19}$, then $g=4+\sqrt{19}$. So our factorization is
$$x^{2}+8x-3=(x+4-\sqrt{19})(x+4+\sqrt{19}).$$
Multiply this out to confirm.

Does that help?
Thanks for the above effort you have put in to explain that, but it's far too advanced for a quadratic that I was trying to understand, where my coursework advises the type to be simple quadratics.

So in my example,

x(x - 3) = 0 this is one solution, and

x = 3 is the second solution.

I think the whole point of my thread was to explain that no matter what anyone does to multiply out brackets as above, you just can't prove that x = 0 unless you are already told that?

so even if you apply ax2 + bx + c = 0

you can say that x2 + 3x = 0

You know that b = 3, and the conclusion is 0, so no maths involved you just have to see that x2 = 0 and 3 (x) must be 3(0) = 0

I can't see how it could ever be proven using algebra?

Kind regards

Casio
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Thanks for the above effort you have put in to explain that, but it's far too advanced for a quadratic that I was trying to understand, where my coursework advises the type to be simple quadratics.

So in my example,

x(x - 3) = 0 this is one solution, and

x = 3 is the second solution.

I think the whole point of my thread was to explain that no matter what anyone does to multiply out brackets as above, you just can't prove that x = 0 unless you are already told that?

so even if you apply ax2 + bx + c = 0

you can say that x2 + 3x = 0

You know that b = 3, and the conclusion is 0, so no maths involved you just have to see that x2 = 0 and 3 (x) must be 3(0) = 0

I can't see how it could ever be proven using algebra?

Kind regards

Casio
Hi Casio, :)

Suppose you have a general quadratic equation of the form, \(ax^2+bx+c=0\mbox{ where }a\neq 0\). Then,

\[ax^2+bx+c=0\]

\[\Rightarrow a\left(x^2+\frac{b}{a}x+\frac{c}{a}\right)=0\]

Since \(a\neq 0\) we have,

\[x^2+\frac{b}{a}x+\frac{c}{a}=0\]

\[\Rightarrow \left(x+\frac{b}{2a}\right)^2-\frac{b^2}{4a^2}+\frac{c}{a}=0\]

\[\Rightarrow \left(x+\frac{b}{2a}\right)^2=\frac{b^2-4ac}{4a^2}\]

\[\Rightarrow x+\frac{b}{2a}=\pm\frac{\sqrt{b^2-4ac}}{2a}\]

\[\Rightarrow x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

So the roots of a quadratic equation is given by,

\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]

In your case you have, \(a=1,\,b=-3\mbox{ and }c=0\). Therefore the roots are,

\[x=\frac{-(-3)\pm\sqrt{(-3)^2-(2\times 1\times 0)}}{2\times 1}\]

\[\Rightarrow x=3\mbox{ or }x=0\]

Hope this clarifies things for you. :)

Kind Regards,
Sudharaka.
 
Jul 22, 2012
35
Thanks for the above effort you have put in to explain that, but it's far too advanced for a quadratic that I was trying to understand, where my coursework advises the type to be simple quadratics.

So in my example,

x(x - 3) = 0 this is one solution, and

x = 3 is the second solution.

I think the whole point of my thread was to explain that no matter what anyone does to multiply out brackets as above, you just can't prove that x = 0 unless you are already told that?

so even if you apply ax2 + bx + c = 0

you can say that x2 + 3x = 0

You know that b = 3, and the conclusion is 0, so no maths involved you just have to see that x2 = 0 and 3 (x) must be 3(0) = 0

I can't see how it could ever be proven using algebra?

Kind regards

Casio
I think you are bit confused here. You're asked to solve x(x-3) = 0. Think about it this way: when do you get 0 from the product of two numbers? When at least one of them is zero! Here you have a product of x and x-3 -- that's x(x-3). You're told that it's zero. It can only be so if x = 0 or x-3 = 0; that's, if x = 0 or x = 3. The fact that $p\times q = 0$ implies that either $p = 0$ or $q = 0$ is something that you should have already been comfortable with. That's probably what you're missing.