Quadratic Equations help/hints

[jillsandwich]

I'm really bad at maths and my teacher isn't answering my emails since he's absent any help would be appreciated, I'm stuck on pretty much everything to be honest but i would appreciate an explanation on the formula. Also sorry if this is formatted wrong I don't really know how to use these sites and I'm slow

 

[Dr.Peterson]

I'm really bad at maths and my teacher isn't answering my emails since he's absent any help would be appreciated, I'm stuck on pretty much everything to be honest but i would appreciate an explanation on the formula. Also sorry if this is formatted wrong I don't really know how to use these sites and I'm slow

Here are some explanations of how to use the quadratic formula:

Quadratic Equations

An example of a Quadratic Equation ... The function makes nice curves like this one

www.mathsisfun.com

The Quadratic Formula Explained | Purplemath

Demonstrates the use of the Quadratic Formula and compares the Quadratic Formula to the solutions found by factoring.

www.purplemath.com

Solving Quadratics by the Quadratic Formula

Easily solve Quadratic equations of the form ax^2+bx+c=0 with step-by-step solutions using the Quadratic Formula.

www.chilimath.com

Once you have the basic idea, you'll probably need some more help; if you show us your work, we can help a lot more than when you don't.

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Welcome to our tutoring boards! :) This page summarizes the main points from our posting guidelines. As our name implies, we provide math help (primarily to students with homework). We do not generally post immediate answers or step-by-step solutions. We don't do your homework. We prefer to...

www.freemathhelp.com

 

[blamocur]

I'm stuck on pretty much everything

I'd suggest working on one thing at a time: have you tried #1? What did you get?

 

[Steven G]

All quadratic equations can be written as ax² + bx + c. These a, b and c are the values which you put into the given formula.

 

[The Highlander]

I'm really bad at maths and my teacher isn't answering my emails since he's absent any help would be appreciated, I'm stuck on pretty much everything to be honest but i would appreciate an explanation on the formula. Also sorry if this is formatted wrong I don't really know how to use these sites and I'm slow

Hi @jillsandwich,

It would appear that you are working (solely) on the use of the quadratic formula here but have no previous experience in its use, hence your difficulties?

I presume that, since you are now learning about the use of this formula, you have already learned about factorising quadratic equations, to find their roots (value(s) of x where \(\displaystyle f(x) = 0\)), and that you already understand that and can do it.

However, many quadratic equations cannot be factorised and that is where the formula may be used to find the root(s) of the equation (if they exist).

It may be helpful to show you an example of using the formula to find the roots of an equation so I will give you a worked example and point out a couple of things that I believe are important to tell students when learning about this topic.

You have been given links to some very helpful websites (which you should visit & study) but I would like to tell you about things that may not be mentioned in them; these are points that I always stress when I am teaching this topic to my students.

Firstly, you have been told that (to use the formula) the equation must be in the form: \(\displaystyle ax^2 + bx + c\). So we have a function, \(\displaystyle f(x)\), where \(\displaystyle f(x) = ax^2+bx+c\text{ (or }y = ax^2+bx+c)\) and we wish to find its root(s).

That is, to find the value(s) of x when \(\displaystyle f(x) = 0\) (ie: y = 0).

The graph of a quadratic function is a parabola () and if we can find the value(s) of x where y = 0 then that will tell us where the graph touches (or crosses) the x-axis (though in some cases the parabolas don't go near the x-axis at all).

So, if we are interested in finding \(\displaystyle f(x) = 0\) (ie: y = 0) then we must set \(\displaystyle ax^2+bx+c\) equal to zero, therefore, what we want is to find the value(s) of x when \(\displaystyle ax^2+bx+c=0\).

I will now show you how to deal with this example: \(\displaystyle x^2-5x=-6\).

Firstly, it's not in the form we want (\(\displaystyle ax^2+bx+c=0\)), so we must re-arrange it by adding 6 to each side to get: \(\displaystyle x^2-5x+6=0\).

I have chosen this particular equation because it is factorisable. I would hope, therefore, that you would be able to say that:-

\(\displaystyle x^2-5x+6=0\implies (x-3)\times (x-2)=0\implies (x-3)=0\text{ or } (x-2)=0\implies x=3\text{ or }x=2\).​

Now here are the important points I wish to draw to your attention:-

When the quadratic expression is in the form: \(\displaystyle ax^2+bx+c=0\), then it is a sum of terms (and each of those terms is an x-term). When I say it is a sum of terms, I want you to remember that, ie: to remember that it is three (x) terms added together (linked by plus (+) signs).

"Whoa!", you say...
"Your 'example' (\(\displaystyle x^2-5x+6=0\)) has a minus sign in it and only two x-terms!"

Yes, I know that is how it looks, but, and this is important (especially when using the quadratic formula), you must be able to recognize that it is actually a sum (addition of) three x-terms each of which has its own (signed) multiplier or coefficient (which may or may not be immediately visible)!

When we put the quadratic into the form: \(\displaystyle ax^2+bx+c=0\), then this is what that really is:-

\(\displaystyle ax^2 + bx^1 + cx^0\)
(where \(\displaystyle a,\, b\And c\) can be +ve or -ve!)

and that is how you should always think of these expressions. (Not thinking of them that way is what leads to many of the mistakes that students make when using the quadratic formula.)

Let me explain further...
The \(\displaystyle a,\, b\And c\) here are, in turn, the coefficients (multipliers) of the three x-terms:-

\(\displaystyle a\) is the coefficient of the \(\displaystyle x^2\) (x squared) term,
\(\displaystyle b\) is the coefficient of the \(\displaystyle x^1\) term (NB: \(\displaystyle x^1\) is just x),
and
\(\displaystyle c\) is the coefficient of the \(\displaystyle x^0\) term (NB: \(\displaystyle x^0=1\) so we get \(\displaystyle c\times 1\) which is just \(\displaystyle c\))

So you really need to start thinking of a quadratic as:-

\(\displaystyle (\pm a)x^2 + (\pm b)x^1 + (\pm c)x^0\)

or (more simply) as...

\(\displaystyle (\pm a)x^2 + (\pm b)x + (\pm c)\)

Now a quadratic must always have an \(\displaystyle x^2\) term. It is because it has an x squared term in it that it is called a quadratic. (A square is a quadrilateral.)

If there is no (apparent) \(\displaystyle a\) in front of the \(\displaystyle x^2\) term then the \(\displaystyle a\) is actually just 1 (ie: you have \(\displaystyle 1x^2\)).

The \(\displaystyle a\) (the \(\displaystyle x^2\) coefficient) may be negative but it cannot be zero (because there must be an \(\displaystyle x^2\) term for the equation to be a quadratic).

However, the coefficients of the other two x-terms (\(\displaystyle x^1\And x^0\)) may be positive or negative or zero!

For example, given the expression: x² - 4 = 0, to use the quadratic formula on it, you should think of it as:-

(+1)x² + 0x¹ + (-4)x⁰ = 0. (or 1x² + 0x + (-4) = 0 will suffice)

It's important to think of quadratics this way to make sure that you get the correct values for the coefficients (\(\displaystyle a,\, b\text{ and }c\)) to plug into the formula.

So, lets now deal with my example...

x² - 5x = -6 ⇒ x² - 5x + 6 = 0

and (to follow my advice) you need to think of that as:-

+1x² + (-5)x¹ + (+6)x⁰ = 0

or, just

1x² + (-5)x + 6 = 0

Now we can clearly see that \(\displaystyle a = 1,\, b = (-5)\And c = 6\).

Our formula is given as:-

[math]x=\frac{-b\pm \sqrt{{\color{Red}b^2-4ac}}}{2a}[/math]
so we can now substitute the (correct) values for \(\displaystyle a,\, b\And c\) into this to get:-

\(\displaystyle \quad \, \quad \, x=\frac{-(-5)\pm \sqrt{{\color{Red}(-5)^2-(4\times 1\times 6)}}}{2\times 1}\\ \,\\ \implies x=\frac{5\pm \sqrt{{\color{Red}25-24}}}{2}\\ \,\\ \implies x=\frac{5\pm \sqrt{{\color{Red}1}}}{2}\\ \,\\ \implies x=\frac{5\pm 1}{2}\\ \, \\ \implies x=\frac{6}{2}\text{ or }x=\frac{4}{2}\\ \,\\ \implies x=3\text{ or }x=2\)

Which gives us exactly the same values for x as we got by factorising the quadratic (up above)! So the formula 'works' (as long as you put in the correct values for \(\displaystyle a,\, b\And c\)).

Let's also have a look at that other expression I mentioned (x² - 4 = 0).

It is a difference of two squares so you should know that:-

x² - 4 = 0 ⇒ (x - 2)(x + 2) = 0 ⇒ x = +2 or (-2).

Alternatively we could say that x² - 4 = 0 ⇒ x² = 4 ⇒ x = +2 or (-2); giving the same result.

But to use the formula on it, I warned that you had to visualize it as:-

(+1)x² + 0x¹ + (-4)x⁰ = 0.

or

1x² + 0x + (-4) = 0

When you see it that way then you can (correctly) identify \(\displaystyle a,\, b\text{ and }c\) as: \(\displaystyle a = 1,\, b = 0\text{ and }c = (-4)\).

Plugging those values into the formula, you get:-

\(\displaystyle \quad \, \quad \, x=\frac{-0\pm \sqrt{{\color{Red}0^2-(4\times 1\times (-4))}}}{2\times 1}\\ \,\\ \implies x=\frac{-0\pm \sqrt{{\color{Red}0-(-16)}}}{2}\\ \,\\ \implies x=\frac{-0\pm \sqrt{{\color{Red}16}}}{2}\\ \,\\ \implies x=\frac{-0\pm 4}{2}\\ \, \\ \implies x=\frac{4}{2}\text{ or }x=\frac{-4}{2}\\ \,\\ \implies x=2\text{ or }x=(-2)\)

So, again, the formula 'works' to give us the correct value(s) for x, ie: the same as we would get by the other (simpler) methods above.

I have gone to the bother of telling you all this because, when this quadratic formula is used, most of the mistakes I see (and occasionally make myself ) is the failure to identify the correct values for the coefficients (\(\displaystyle a,\, b\And c\)) to put into the formula and then (just as often) the incorrect handling of the +ve and -ve numbers once they are substituted into it.

You need to be sure that you identify \(\displaystyle a,\, b\And c\) correctly in the first instance (following my advice above) and then take the greatest of care handling the signs (+ & -) properly when carrying out the arithmetic to evaluate the final answer(s) that the formula may produce.

PS: If you are somewhat daunted by the final section of your exercise (Questions 16 - 21) then you may find this useful:-

Q.16 Gives you: \(\displaystyle x+\frac{2}{x}=7\) which you must rearrange into the form ax² + bx + c = 0.

If you start by multiplying both sides by x then you can then more easily rearrange it to get:-

\(\displaystyle \quad \, \quad \, x+\frac{2}{x}=7\\ \,\\ \implies x\left(x+\frac{2}{x}\right)=7x\\ \,\\ \implies x^2+\frac{2x}{x}=7x\\ \,\\ \implies x^2+2=7x\\ \,\\ \implies x^2-7x+2=0\)

Hope that helps.

 

[The Highlander]

Postscript:

On The Discriminant.

You may have noticed that, in the post above, I put several words into (possible) plurals by adding an 's' (in brackets) at the end; this is to indicate that the word might be singular or plural. Why?

Well, some quadratics have two roots (the graph crosses the x-axis at two different points) and some have only one root (the graph just touches the x-axis at only one point) but there are also some quadratics that have no (real) roots (the graph doesn't touch the x-axis at all; it is either completely above or below the axis).

So there may be one, two or no roots at all to a quadratic equation.

You may also have noticed that I coloured everything under the square root red in the formula:-

[math]x=\frac{-b\pm \sqrt{{\color{Red}b^2-4ac}}}{2a}[/math]
That is because that part of the formula (\(\displaystyle b^2-4ac\)) has a special importance.

It is called the Discriminant and, once you have mastered the use of the formula, that is, almost certainly, the next topic that you will be expected to learn about.

Why is it so significant? Well, if \(\displaystyle b^2-4ac\) is a positive number then it will be possible to find a square root for it. That means the formula will give rise to two answers (because you will add \(\displaystyle \sqrt{b^2-4ac}\text{ to}-b\) to get one answer and then subtract \(\displaystyle \sqrt{b^2-4ac}\text{ from}-b\) to get a second answer. So, if \(\displaystyle b^2-4ac\) is a positive number then you can tell (right away) that the graph of the quadratic will cross the x-axis at two different points.

If, however, \(\displaystyle b^2-4ac\) is zero (0) then the square root of 0 is just 0 and adding or subtracting 0 to -b won't affect it and so the formula just becomes \(\displaystyle \frac{-b}{2a}\) which is just a single result so the quadratic only has one, single root which means the graph just touches the x-axis at only one point.

And, finally, if \(\displaystyle b^2-4ac\) is a negative number, then there is no square root for a negative number!
You cannot find any number that when multiplied by itself produces a negative result (eg: 2 × 2 = 4 and (-2) × (-2) = 4). So if \(\displaystyle b^2-4ac\) is negative, then it becomes impossible to evaluate a result for the formula and the quadratic therefore has no roots which means the graph doesn't touch the x-axis at all; it is either completely above or below the axis.

Now, in more advanced Mathematics, we do come up with a way of finding square roots for negative numbers. We just say: let's imagine that there is a number (we call it "\(\displaystyle i\)") that when it is multiplied by itself the answer is (-1), ie: \(\displaystyle i^2=-1\) and that produces a whole new set of numbers called Imaginary Numbers (as opposed to Real Numbers).

That then allows us to find imaginary roots to quadratics where \(\displaystyle b^2-4ac\) is negative and we can say then, that there exist imaginary roots. (Though, we usually just say that a quadratic has no real roots if \(\displaystyle b^2-4ac\) is negative.)

But all of that is still in the distant future for you (if you continue your journey into Mathematics).

Good luck with your exercise, @jillsandwich.

Please come back and show us your work and if there are any errors (or you are still having difficulties) further advice will be offered.