Question

Tutorial post: Finding the REAL roots/zeros/x-intercepts (let’s pass on explaining imaginary numbers for now) of a quadratic equation because so many people need help with that unfortunately. I’ll first write out three ways to find them, then dive into each one after
P.S. please refrain from commenting until I’m done, and yes, I pre wrote this, I’m not spending 10 hours on the spot writing this 😔🤧

Answer 1

The 3 ways to find the roots are:
Factoring- either by regular factoring or factor by grouping
The quadratic formula
Graphing it

Answer 2

The factoring method is quite commonly used to easily find the real roots of a quadratic equation

2 things to know before doing this is
Firstly, when we have \[ (x-a)(x-b) \], we set the other side to 0, because we want the x values where y is 0, so we have (x-a)(x-b) =0. And because of this, we separate it into two equations, where x-a=0 and x-b=0, so at the end, positive a and positive b are the x-intercepts.
Secondly, the FOIL method used to check.
FOIL method: forward, outside, inside, last
(a+b)(c+d) = ac +ad + bc + bd (usually we are able to combine ad and bc)



And how do we get to factored form from standard form?
Well, let’s first take an example: \[ x^2 + 7x + 6 \]
In this, from seeing x^2, we know that we can set it to (x +/- _)(x +/- _)
And now we want to set the signs, we can set this one right when we see the equation in this example, because both b and c are positive integers (in ax^2 + bx + c)
So we can write (x+_)(x+_)
Nowe we have to fill in the two blanks. We look at the c value, and see that it is 6, from this, we can list out all the factors that multiply to 6, and then find which pair adds up to 7
So we would have:
1 & 6 -> 1+6 = 7
2 & 3 -> 2+3=5
So, from above, we can see that the two blanks we are about to fit in are 1 and 6. So we have \[ (x+6)(x+1) \] and you can also check this by using the FOIL method

Check: (x+6)(x+1) = x^2 + x + 6x + 6 = x^2 + 6x + 6
So we are correct

And from (x+6)(x+1) we can determine that -6 and -1 are the roots, and so the coordinates are (-6,0) and (-1,0)





And so now, we can move on to the next scenario, if we had, for example \[ x^2 -7x + 6 \] As you can see, I’m using the same numbers, just changing b into a negative integer
From this, we know that we need to integers that multiply up to 6, but add up to -7, in these kind of cases, it is obvious that the two signs in the the expression are both negative, because negative and negative makes positive, which gets our c value to be a positive integer, but also gets our b value to a negative number since -x+(-y) = -x-y, so it is negative
So we can set this up to (x-_)(x-_)
From here, we can see that again, 1 times 6 is six, and adds up to 7, so we have \[ (x-6)(x-1) \] And again, we can check using the FOIL method

Check: (x-6)(x-1) = x^2 -x - 6x + 6 = x^2 -7x + 6

And from (x-6)(x-1) we can determine that 6 and 1 are the roots, so the coordinates are (6,0) and (1,0)





The next scenario is where one sign is positive, the other is negative
For example, if we had \[ x^2 +2x - 15 \] So from this, we can go straight on into finding the factors of -15 that add up to 2
-1 & 15
-15 & 1
-3 & 5
-5 & 3
From here, you can see that only -3 and 5 add up to 2, so our blank spaces are -3 and 2, so we have \[ (x-3)(x+2) \] so the roots are 3 and 2, the coordinates are (3,0) and (-2,0)
Also, something you should observe is, you need two numbers… one positive one negative… but your b value has to be positive, this obviously means that the number that is positive must be larger than the absolute value of the negative integer.

Another example is \[ x^2 -4x -21 \]
Listing out the factors,
-1, 21
-21, 1
-3, 7
-7, 3

Obviously, we choose either form -3, 7 and -7, 3 and knowing that since b is negative, we need the negative absolute value integer bigger than the positive one, so we choose -7, 3, since -7+3 is -4, which is just the b we need. So we fill in the blank spaces and get \[ (x-7)(x+3) \] and from there, we can determine that the roots are 7 and -3, so the coordinates are (7,0) and (-3,0)

Answer 3

The Quadratic formula is

\[\frac{ -b \pm \sqrt{b^2 -4ac} }{ 2a }\] where \[ ax^2 + bx +c \]


Let’s first use an easy example, with integers as its roots

Find the roots of: \[ x^2 -5x + 6 \] So we know that
a is 1
b is -5
c is 6
Plug it into the equation \[ \frac{ 5 \pm \sqrt{25 -24} }{ 2 } = \frac{5 \pm \sqrt{1} }{2} \] From here, we know that it is \[ \frac{ 5 \pm 1 }{2} \], and then we can separate this to two equations, \[ \frac { 5 + 1}{ 2 } \] and \[ \frac {5-1}{2} \]
At the end, this gets us the values of 3 and 2, which are the x intercepts, so we have (3,0) and (2,0)


Now let’s a do a non-integer solution one
Find the roots of: \[ x^2 -6x +6 \] so we have
a = 1
b = -6
c = 6
So then we get
\[\frac{ 6 \pm \sqrt{36 -24} }{ 2 } = \frac{ 6 \pm \sqrt{12} }{2 }\] Then we can simplify it to \[ \frac { 6 \pm \sqrt{12} }{ 2 } \] We can separate this into two equations and also simplify the square root so we have
\[ \frac { 6 + 2\sqrt{3} }{ 2 } \] and \[ \frac {6 - 2\sqrt{3} } {2} \] so we can simplify it get it to \[ 3 + \sqrt{3} \] and \[ 3 - \sqrt{3} \] And so these would be your exact answers. And if you wanted to put this into factored form, we would write \[ x - 3- \sqrt{3})(x-3+ \sqrt{3}) \]And to find the roots of this function in decimal form/approximate answer, simplify put the two into a calculator, and you’ll get approximately 1.268 and 4.73. An easy way to check if you are approximately correct is graphing this, which I will talk about next.

Answer 4

Graphing method: This is used the most when either we have integer solutions but are too lazy to find them, or we use this to find the approximate decimal form of the solutions, also because we are too lazy to use the above two methods. But I suggest using this method if you are familiar with the above two already, because it’ll save a lot of time.
So you can go to Desmos or GeoGebra and simplify type in the equation you need to find the roots of, look at the graph, and find the points where the parabola crosses the x axis. Those are your coordinates.

Answer 5

If you have questions/comments feel free to message me, I’ll reply when I can…

Answer 6

@Tranquility request post lock and deletion of side comments o-O