Quadratic Equations
(equations with x2)
This unit will teach you how to solve
equations like
2x2 – 7x – 3 = 0
a2 – 9 = 0
2b2 + 5b = 0
x2 – 5x = 6 etc…
But first….we need two bits of
BACKGROUND KNOWLEDGE!!
Background #1
20= 0
20= 0
05= 0
a0= 0
If we multiply any two numbers together and
the result is zero, then…..
One of the numbers must be zero! (Maybe
both are)
Now, using letters rather than numbers,
If a  b = 0 then we can conclude….
Either a = 0 OR b = 0
If (a – 2 )  (a – 1) = 0 then we can conclude….
Either a – 2 = 0 OR a – 1 = 0
Making a the
subject of
a=2
a = 1 each equation.
Background #2
You need to revise how to factorise quadratics.
Study THE FORMATS in this table:
TYPE
Distinguishing Features
Example
Answer
1
• 2 terms each with x
• common factor
x2 – 5x
x(x – 5)
2
•2 terms, only one x
• both terms are squares
• always a minus sign
x2 – 4
(x – 2)(x + 2)
3
•3 terms: x2, x and a
constant (number)
x2 – 3x – 28
(x – 7)(x + 4)
4
•3 terms like Type 3
• Number in front of x2
SHOW ME
2x2 – 7x – 4
SHOW ME
(2x + 1)(x – 4)
Example #1
Solve x2 – 16 = 0
STEP 1
Use Background 2 information to
classify x2 – 16 as a TYPE 2 & factorise
to (x – 4)(x + 4)
STEP 2
Rewrite the original equation as
(x – 4)(x + 4) = 0
STEP 3
Noting that this really means
STEP 4
(x – 4)  (x + 4) = 0
we can now use Background 1 knowledge
to conclude that
x–4=0
OR
x+4=0
Make x the subject of each
x=4
OR
x=–4
Example #2
Solve x2 + 7x = 0
STEP 1
Use Background 2 information to
classify x2 + 7x as a TYPE 1 & factorise
to x ( x + 7)
STEP 2
Rewrite the original equation as
x(x + 7) = 0
STEP 3
Noting that this really means
STEP 4
x  (x + 7) = 0
we can now use Background 1 knowledge
to conclude that
x =0
OR
x+7=0
Make x the subject of the 2nd one
x =0
OR
x=–7
Example #3
Solve x2 + 4x + 3 = 0
STEP 1
Use Background 2 information to
classify x2 + 4x + 3 as a TYPE 3 &
factorise to (x + 1) ( x + 3)
STEP 2
Rewrite the original equation as
(x + 1)(x + 3) = 0
STEP 3
As this means
(x + 1)  (x + 3) = 0
we can conclude that
x+1=0
STEP 4
OR
x+3=0
Make x the subject
x = –1
OR
x=–3
Example #4
Solve 2x2 –7x + 3 = 0
STEP 1
Use Background 2 information to
classify 2x2 – 7x + 3 as a TYPE 4 &
factorise to (2x – 1 ) ( x – 3) SHOW ME
STEP 2
Rewrite the original equation as
(2x – 1 )(x – 3 ) = 0
STEP 3
2x – 1 = 0
STEP 4
Make x the subject
x =½
OR
OR
x–3 =0
x=3
Example #5
STEP 1
Solve 2x2 – 50 = 0
Divide through by 2.
Becomes x2 – 25 = 0
See note
Now this is a TYPE 2 & factorises to
(x – 5)(x + 5)
STEP 2
Rewrite as
(x – 5 )(x + 5 ) = 0
STEP 3
x–5 =0
OR
STEP 4
Make x the subject
x =5
OR
x+5 =0
x=–5
Example #6
STEP 1
SEE NOTE
Solve 3x2 – 9x – 30 = 0
Divide through by 3. x2 – 3x – 10 = 0
x2 – 3x – 10 is now a TYPE 3 &
factorises to (x – 5)(x + 2) SHOW ME
STEP 2
Rewrite as
(x – 5 )(x + 2 ) = 0
STEP 3
x–5 =0
OR
STEP 4
Make x the subject
x =5
OR
x+2 =0
x=–2
Example #7
STEP 1
STEP 2
Solve 2x2 – 7x = 4
Regroup to get 0 on right hand side. See note
Becomes 2x2 – 7x – 4 = 0. Do as Type 4
SHOW ME
Rewrite the original equation as
(2x + 1 )(x – 4 ) = 0
STEP 3
Noting that this really means
STEP 4
Make x the subject
(2x + 1)  (x – 4 ) = 0
we can now conclude that
2x + 1 = 0 OR
x–4 =0
x = -½
OR
x=4
21
Example #8 Solve x   4
x
STEP 1 First, KILL THE FRACTION by
multiplying throughout by x PLEASE EXPLAIN
STEP 2
STEP 3
STEP 4
x2 = 21 – 4x
Second, get 0 on the right hand side:
x2 + 4x – 21 = 0
Do as a Type 3:
(x + 7)(x – 3 ) = 0
x + 7 = 0 OR x – 3 = 0
x=–7
OR x = 3
PROBLEM SOLVING !!
Example #9
Two numbers have a sum of 12 and a product of
35. Find the numbers.
With these questions, always begin by saying
“Let one number be equal to x”.
Since they both add to 12, that means the other
number must be 12 – x. So we can also say
“Let the other number be equal to 12 – x”.
Now we’ll put this into symbols
As our numbers are x and 12 – x, and we know their
product is 35, we can say
x (12 – x) = 35
Expanding,
12x – x2 = 35
Rearranging & changing all signs,
x2 – 12x + 35 = 0
Remember to
make sure you
answer the
question!
Factorising as a Type 3,
(x – 5)(x – 7) = 0
x = 5 or x = 7
FINAL ANSWER: The numbers are
5 and 7.
Solving Quadratics using the Graphics.....
The background theory goes like this.....
Suppose you have any equation with one unknown (x).
The format will always be like this:
LEFT HAND SIDE = RIGHT HAND SIDE
On your graphics, by hitting Y= and then entering
Y1 = LEFT HAND SIDE
Y2 = RIGHT HAND SIDE
graphing and using 2nd TRACE INTERSECT,
you can find where they meet, and this is the
solution/s to the equation!!
Solving Quadratics using the Graphics.....
Example 1. Solve x2 – 3x – 4 = 0
STEP 1 Hit Y= and enter Y1 = x2 – 3x – 4 and
Y2 = 0 (the x-axis)
STEP 2 Hit WINDOW and
make XMIN – 5, XMAX 5
STEP 3 Hit ZOOM. Choose Option 0
sure you can see where the
(ZOOMFIT) Make
graphs meet!
Y1
Y2
STEP 4 Hit 2nd TRACE. Choose
Option 5 (INTERSECT) &
hit ENTER 3 times
Doing this twice, with the cursor positioned near each intersection
point, will yield the results x
= – 1 and x = 4
Solving Quadratics using the Graphics.....
Example 2. Solve 2x2 – 5x = 3
STEP 1 Hit Y= and enter Y1 = 2x2 – 5x and
Y2 = 3
STEP 2 Hit WINDOW and
make XMIN – 5, XMAX 5
STEP 3 Hit ZOOM. Choose Option 0
sure you can see where the
(ZOOMFIT) Make
graphs meet!
STEP 4 Hit 2nd TRACE. Choose
Option 5 (INTERSECT) &
hit ENTER 3 times
Doing this twice, with the cursor positioned near each intersection
point, will yield the results x
= – 0.5 and x = 3
QUICK QUIZ!!
(1) The solutions to x2 – 16 = 0 are:
x = 4 or 0
x = – 4 or 0
x = 4 or – 4
x = 16 or –16
(2) The solutions to x2 – 4x = 0 are:
0 and – 4
4 and – 4
2 and – 2
0 and 4
(3) The solutions to x2 – x = 12 are:
4 and – 3
3 and – 4
2 and – 6
0 and 3
(4) The solutions to 25 – x2 = 0 are
5 and – 5
5 and 0
– 5 and 0
5 and 5
(5) The solutions to 3x2 + x = 10 are
- 5/3 and 2
5/3 and – 2
5/3 and 2
- 5/3 and – 2
(6) To solve 2x2 – 5x = 3 on the graphics you
could: (there are TWO correct answers)
Draw Y1 = 2X2 – 5X – 3 and see where
it cuts the y-axis
Draw Y1 = 2X2 – 5X – 3 and see where
it cuts the x-axis
Draw Y1 = 2X2 – 5X and Y2 = 3,
and see where they intersect
Draw Y1 = 2X2 – 5X and Y2 = 3,
and see where they cut the x-axis
Factorising trinomials (Type 4) – example
Factorise 2a2 – 5a + 3
STEP 1
STEP 2
Mult. 2 by 3 to get 6
Set up brackets (2a.......)(2a.......)
2
Find two numbers that
MULTIPLY to make
ADD to make
+6
–5
Numbers are – 3 and – 2
STEP 3
STEP 4
Insert numbers into brackets in Step 1
(2a  3)(2a  2)
2
Divide denominator (2) fully into 2nd bracket
Ans (2a – 3)(a – 1)
NOTE about Type 3 & 4
Sometimes you will get a quadratic (like in
Example 5 or 6) which might LOOK LIKE A
TYPE 4 (it has a number in front of the x2). But on
looking closer, you notice you can divide through
by a number, making it into a TYPE 3 or 2.
Because Type 3s & Type 2s are easier to factorise
than Type 4s, this makes good sense!
Examples such as 3x2 – 9x – 30 can also be done as
Type 4s, but it takes more work!!
BACK
Factorise 2x2 – 7x – 4 (Example 7)
STEP 1
STEP 2
Set up brackets (2 x.......)(2 x.......)
2
Find two numbers that
MULTIPLY to make
ADD to make
–8
–7
Numbers are – 8 and + 1
STEP 3
STEP 4
Insert numbers into brackets in Step 1
(2 x  8)(2 x  1)
2
Divide denominator (2) fully into 1st bracket
Ans (x – 4)(2x + 1)
When solving quadratics, ALWAYS make sure
you get 0 on the right hand side, and that your
equation has the format
ax2 + bx + c = 0
where a, b, c are constants (numbers).
BACK
Brilliant!!
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
Stiff cheddar! Have
another go
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