Grade 10 Quadratic Unit

The Basics of the Algebraic portion


Quadratics are an important math unit that is vital to engineering and certain fields of business. The unit covers the science of the parabola and its teachings will ensure the preservation of structures such as the decadent roman Colosseum(picture on the left) and the current works of architecture that we rely on today.


When we factor we actually do the opposite of expanding as we try to shrink the equation/expression.When every term of a polynomial is divisble by the same constant then the constant is called a common fact.


In this equation a is the common factor

Polynomials are not considered to be fully factored until the greatest common factor(G.C.F) has been factored out

Let us consider the following expression, 8x-12y

Now both 8x and 12y can be divided by 4 as 4 is thre G.C.F.

we would therefore start our factored expression off with 4 and then continue the dividing process

8x divided by 4 would be 2x and 12y divided by 4 would be 3y. We now have the information needed to make our factored form.

Therefore the new equation would be


Please view the example as explained by Khan Academy below. The first 4 minutes will suffice in explaining the very basics of factoring.

Factoring Quadratic Expressions


When we multiply binomials we expand with the distributive property. What this means is that we take each term in the bracket and multiply it by the terms in the bracket next to it.

Let us consider the following:


if we expand this by multiplying each term in the first bracket by the 2nd we get:

(x)(x)+ bx+ax+ab

After we expand our next step is to to simplify.

x^2 +(a+b)x+ab

If we were to apply the distributive property to an exponent it would work like this in theory.


we would first have to make 2 groups as (x-5) is all to the exponent of 2.

Then we would get (x-5)(x-5) and then work on expanding from there.

With that accomplished we should get x^2 -5x-5x+25

After simplifying(adding the x's)

we will get x^2-10x+25

Therefore x^2-10x+25 would be our final answer.


The simplest way to solve for the value of x is to use the quadratic formula(as seen on the picture to the left). It is used to solve questions in the form of ax^2 + bx + c= 0. For the purpose of explanation we will solve 1x^2 +3x-4=0 using the quadratic formula.

Our first step will be to write out the formula of x= -b+/- √b^2-4ac divided over 2a.

Our next job is to start substituting the values we received from the equation. In this example, 1 would be a, 3 would be b and -4 would be c. when we substitute these values into the formula we get the following: -3 +/- √ 3^2-4(1)(-4) all divided by 2(1)

Most of the work will come from the former step as the only thing left to do is to use the BEDMAS order(brackets,exponents,multiply,addition,subtraction) to break it down into our 2 x values.

With that said it will break down to -3 +/- √ 9+16 all divided by 2. The 9 came from solving the exponents the 2nd part of the BEDMAS order

when we add 9 and 16 the equation will become x=-3+√25 all divided by 2

From here we find the square root of 25 before we proceed further. The square root of 25 is 5 therefore √25=5

Now because of the + and - sign just after the b value we have 2 answers

one will be positive and the other negative

x= (-3+5)/2 and (-3-5)/2

we must solve the brackets and divide by 2 to receive our two answers

They will be x=-4 and x=1

Therefore the x intercepts will be -4 and 1

Completing the Square

Consider the following equation: y=x^2+6x+11

The equation is currently in standard form, if we wished to find the vertex form(so we could figure out how to graph it later on) we would have to complete the square

The first step would be to block off the very first 2 terms so we would get


Our Next step would be to factor out the A but since the A value is equal to 1 we may skip this step

Our 3rd step would be to divide the middle term(6x in this example) by 2 and then square it

It would look like (6/2)^2=9

After getting the answer we plug it back into the equation to get y=(x^2+6x+9-9)+11

Lastly we take the negative out of the bracket by multiplying it by the A value outside of the bracket(1)


After taking out the negative we square the equation to get