Am I doing these problems correctly?

**uperkurk** · October 20th 2012, 09:19 PM

$x^2+7x+10=(x+2)(x+5)$
$x^2+5x+6=(x+2)(x+3)$
$x^2-8x+15=(x-3)(x-5)$
$x^2-x-30=(x-6)(x+5)$

These are ones I made up myself.

$x^2+x-2=(x+2)(x-1)$
$x^2-4-8=(x+2)(x-8)$
$x^2+3+9=(x+3)(x+3)$
$x^2+3-9=(x-3)(x-3)$

$x^2-4x+7$ This can not be solved right? As the only factor of 7 is 1 and 7, which I can not get -4 or 4 from? Are the others correct? And also does it matter if I have

$x^2-4-8=(x+2)(x-8)$ or $x^2-4-8=(x-8)(x+2)?$

Once a quadratic has be factorised, what is the next step? Can someone explain to be please.

**MarkFL** · October 20th 2012, 09:29 PM

The first 4 you factored correctly.

Of the ones you made up, the first is correct, but the last 3 are not correct, even assuming the middle term has an $x$ as a factor.

As far as which order you write the factors, this does not matter, as per the commutative property of multiplication, i.e $ab=ba$ .

You are right about the last one, it cannot be factored in the traditional sense. In fact, it has complex roots. One way to check a quadratic to see if it can be factored is to see if the discriminant is a perfect square.

Assuming you are trying to solve for $x$ when a quadratic equals zero, once you have the quadratic factored, you then equate each factor to zero and solve for $x$ to find the roots, or values that make the quadratic zero.

If you have the product of some factors being equal to zero, then if any one of the factors is equal to zero, then the entire product is equal to zero. This is called the zero-factor property. If $ab=0$ then we know either $a=0$ or $b=0$ must be true.

**uperkurk** · October 20th 2012, 09:45 PM

I think the reason why the other 3 I made up are wrong is because I chose bad numbers? for example 3 and 9? I never really put too much thought into it when making them up. Anyways if you have time could you explain this bit "you then equate each factor to zero and solve for x to find the roots, or values that make the quadratic zero."

If you could show the steps on one of my problems maybe I can then apply that to the other questions. I haven't really learnt the next steps but depending on how hard it seems I'll head over to youtube

**MarkFL** · October 20th 2012, 09:53 PM

Suppose you are given:

$3x^2-19x-14=0$

We want to find the values of $x$ which make this equation true, that is, we want to solve for $x$ .

We may factor the quadratic and write:

$(3x+2)(x-7)=0$

Now, this equation will be true if:

$3x+2=0$

$x=-\frac{2}{3}$

It will also be true if:

$x-7=0$

$x=7$

Thus, the two roots, or solutions, are:

$x=-\frac{2}{3},\,7$

**uperkurk** · October 20th 2012, 10:01 PM

So every quadratic has 2 solutions? Do they ever have 1 solution? I am following this guide Factoring Quadratics and using the formula. I'll let you know what I get for one of my examples.

**MarkFL** · October 20th 2012, 10:05 PM

Some quadratics have 1 solution, which is considered a repeated root. For example:

$x^2+6x+9=0$

$(x+3)^2=0$

This quadratic then has the repeated root $x=-3$ . If the discriminant is equal to zero, then you will have a repeated root. If it is positive then you will have two distinct real roots, and if it is negative you will have two complex roots.

**uperkurk** · October 20th 2012, 10:25 PM

Hmm this isn't going so well lol.

I have $2x^2+5x+6$ then I use the formula

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

$x=-5 \pm\sqrt\frac{5^2-4\cdot2\cdot6}{2\cdot2}$

$x=-5 \pm\sqrt\frac{25\cdot12-4}{4}$

$x=-5 \pm\sqrt\frac{296}{4}$

$-5 + 17.088=12.088$
$-5 - 17.088=-22.088$

Excuse the slow response I'm also trying to learn latex to make it easier for people helping me. Where am I going wrong?

**MarkFL** · October 20th 2012, 10:29 PM

The quadratic formula is:

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

Your use of $\LaTeX$ is a great way to make it easier to read your work!

edit: your work under the radical in your previous post is also wrong, watch for this when you use the quadratic formula.

**uperkurk** · October 20th 2012, 10:41 PM

Rather than just guessing I'm going to read up on it for a bit because I don't really understand it, just quickly though what do you mean radical?

**MarkFL** · October 20th 2012, 10:52 PM

The radical is the square root symbol. The expression under the radical in the quadratic formula is referred to as the discriminant.

Done correctly, you will find the discriminant is negative in the problem you gave, so your roots will be complex, that is, having real and imaginary parts.

**uperkurk** · October 20th 2012, 11:06 PM

Thanks for the help. I'm going to read up a bit more on this before trying anymore problems

Thanks though

**MarkFL** · October 20th 2012, 11:19 PM

I recommend also looking at the method called completing the square, and hopefully you will see a geometric representation of the technique which helps to show why the technique works. This is also how the quadratic formula is derived. Good luck with your research!

**uperkurk** · October 21st 2012, 07:49 AM

Rather than open a new thread I have another quadratic problem, this time with graphing them.

"Simplify and graph the following quadratic equation"

$x^2+7x+10$

$=(x+2)(x+5)$

Now to graph it.

$a=1, b=-7, c=10$

$x=\frac{-7}{2}=-3.5$

$(-5.5)^2+7(-5.5)+10=y$
$(-2.5)^2+7(-2.5)+10=y$
$(-3.5)^2+7(-3.5)+10=y$
$(-2.5)^2+7(-2.5)+10=y$
$(-1.5)^2+7(-1.5)+10=y$

But when I plot these I get a weird snake type of graph. Where am I going wrong? I even typed those equations exactly into wolfram and it gave me the same answer as I got.

**MarkFL** · October 21st 2012, 08:10 AM

If I were going to sketch the graph, I would write it in vertex form:

$y=x^2+7x+10=\left(x^2+7x+\left(\frac{7}{2} \right)^2 \right)+10-\left(\frac{7}{2} \right)^2=\left(x+\frac{7}{2} \right)^2-\frac{9}{4}$

Now, it is easy to see that we merely need to translate the graph of $y=x^2$ to the left $\frac{7}{2}$ units and $\frac{9}{4}$ units down.

Also, you know the roots are (-5,0) and (-2,0) and the axis of symmetry is the line $x=-\frac{7}{2}$ .

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