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 #12 0 = 0 2 0 =

0 5 =

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)

If a b = 0 then we can conclude….Either a = 0 OR b = 0

Now, using letters rather than numbers,

If (a – 2 ) (a – 1) = 0 then we can conclude….Either a – 2 = 0 OR a – 1 = 0

a 0 = 0

a = 2 a = 1

Making a the subject of each equation.

Background #2You 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 factorx2 – 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

2x2 – 7x – 4 (2x + 1)(x – 4)

SHOW ME

SHOW ME

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

(x – 4) (x + 4) = 0

we can now use Background 1 knowledge to conclude thatx – 4 = 0 OR x + 4 = 0

STEP 4

x = 4 OR x = – 4Make x the subject of each

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

x (x + 7) = 0

we can now use Background 1 knowledge to conclude thatx = 0 OR x + 7 = 0

STEP 4

OR x = – 7Make x the subject of the 2nd one

x = 0

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 OR x + 3 = 0

STEP 4

OR x = – 3Make x the subject

x = –1

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)

STEP 2 Rewrite the original equation as (2x – 1 )(x – 3 ) = 0

STEP 3 2x – 1 = 0 OR x – 3 = 0

STEP 4

OR x = 3

Make x the subject

x = ½

SHOW ME

Example #5 Solve 2x2 – 50 = 0

STEP 1 Becomes x2 – 25 = 0

STEP 2 Rewrite as (x – 5 )(x + 5 ) = 0

STEP 3 x – 5 = 0 OR x + 5 = 0

STEP 4

OR x = – 5 Make x the subject

x = 5

Divide through by 2. See note

Now this is a TYPE 2 & factorises to (x – 5)(x + 5)

Example #6 Solve 3x2 – 9x – 30 = 0

STEP 1

x2 – 3x – 10 is now a TYPE 3 & factorises to (x – 5)(x + 2)

STEP 2 Rewrite as (x – 5 )(x + 2 ) = 0

STEP 3 x – 5 = 0 OR x + 2 = 0

STEP 4

OR x = – 2 Make x the subject

x = 5

Divide through by 3.

SEE NOTE

SHOW ME

x2 – 3x – 10 = 0

Example #7 Solve 2x2 – 7x = 4

STEP 1 Regroup to get 0 on right hand side. Becomes 2x2 – 7x – 4 = 0. Do as Type 4

STEP 2 Rewrite the original equation as (2x + 1 )(x – 4 ) = 0

STEP 3 Noting that this really means

(2x + 1) (x – 4 ) = 0

we can now conclude that

2x + 1 = 0 OR x – 4 = 0

STEP 4

OR x = 4Make x the subject

x = -½

SHOW ME

See note

Example #8 Solve 421

x

x

STEP 1 First, KILL THE FRACTION by multiplying throughout by x

x2 = 21 – 4xSTEP 2 Second, get 0 on the right hand side:

x2 + 4x – 21 = 0

STEP 3 Do as a Type 3:(x + 7)(x – 3 ) = 0

STEP 4 x + 7 = 0 OR x – 3 = 0

x = – 7 OR x = 3

PLEASE EXPLAIN

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 Factorising as a Type 3,

(x – 5)(x – 7) = 0

x = 5 or x = 7FINAL ANSWER: The numbers are 5 and 7.

Remember to make sure you answer the question!

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 (ZOOMFIT)

STEP 4 Hit 2nd TRACE. Choose Option 5 (INTERSECT) & hit ENTER 3 times

Make sure you can see where the graphs meet!

Doing this twice, with the cursor positioned near each intersection

point, will yield the results x = – 1 and x = 4

Y1

Y2

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 (ZOOMFIT)

STEP 4 Hit 2nd TRACE. Choose Option 5 (INTERSECT) & hit ENTER 3 times

Make sure you can see where the graphs meet!

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 = 16 or –16

x = 4 or – 4

x = – 4 or 0

(2) The solutions to x2 – 4x = 0 are:

0 and – 4

0 and 4

2 and – 2

4 and – 4

(3) The solutions to x2 – x = 12 are:

4 and – 3

0 and 3

2 and – 6

3 and – 4

(4) The solutions to 25 – x2 = 0 are

5 and – 5

5 and 5

– 5 and 0

5 and 0

(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 and Y2 = 3,and see where they cut the x-axis

Draw Y1 = 2X2 – 5X and Y2 = 3,and see where they intersect

Draw Y1 = 2X2 – 5X – 3 and see where it cuts the x-axis

Factorising trinomials (Type 4) – example

Factorise 2a2 – 5a + 3

STEP 1 Set up brackets

STEP 2

STEP 3

Find two numbers that

MULTIPLY to make + 6

ADD to make – 5

Numbers are – 3 and – 2

Insert numbers into brackets in Step 1

Ans (2a – 3)(a – 1)

2

.......)2.......)(2( aa

Mult. 2 by 3 to get 6

2

)22)(32( aa

STEP 4 Divide denominator (2) fully into 2nd bracket

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 Set up brackets

STEP 2

STEP 3

Find two numbers that

MULTIPLY to make – 8

ADD to make – 7

Numbers are – 8 and + 1

Insert numbers into brackets in Step 1

Ans (x – 4)(2x + 1)

2

.......)2.......)(2( xx

2

)12)(82( xx

STEP 4 Divide denominator (2) fully into 1st bracket

When solving quadratics, ALWAYS make sure you get 0 on the right hand side, and that your

equation has the format

ax2 + bx + c = 0where a, b, c are constants (numbers).

BACK

Brilliant!!

BACK

Stiff cheddar! Have

another go

BACK