Lecture notes

Quadratic Equations
Def : An
-
equation of the form aX2tbxtc=0 where
a. b. & C are real ribs L w/ a 0) is a
"
quadratic equation =


[ form
"
in standard .




-
A quadratic equation will have :


* 2 solutions
lil if b? Uac > 0 ( Xi & ✗ a) ,



*
Lii) ( Xi
1 solution ✗ 2=4) if 62 Hae so = -




,


* Liii ) no solutions if 62 Uac < 0 ,
-




-
There are 3
major techiqeus for solving quadratic equations :
2
i
factoring
Quadratic formula } ( Most common)


3
Square
-


root both sides ( special case)
-
for 1
factoring Quad form , the quadratic
& 2
equation
must be in standard form : aX2t6✗tC=O

factoring
1


-
We make use of the following FACT if A. 8=0 then ,


A- =D or 8=0 (or both )
in order to solve
eq a✗ʰtbXtc -0
a quad , by factoring
the
left -
hand side &
setting the factors = to
,
0 to solve fork
the equation ✗21-7 ✗
-

Eg : Solve -10 =


first, ✗21-7×1-10=0
→
put it in standard form :

Then
, factor the LHS (✗ 1- 2) 1×+51=0 LHS : Left
This implies that
{ Kt 2=0 ✗ t 5 = 0 ✗
✗ =


=
-2
-5
hand side

Solve ✗2=4×-4
-




Eg :

✗2- 4×+4=0 ( binomial squared )
1×-211×-21=0 or 1×-212=0

so
,
{ ✗ -2=0 ✗ 2 =

, ✗ 21-1=0 V. 8 : ✗ 2=-1
Eg No Solve
solution
:

5×2 = Fi
-




for ✗ 21-1=0 ✗ = if
a = 1
,
6=0 ,
Csl
so 62 Uac so -44 )(1)
-


= -4<0

Eg : Solve the equation ✗ 3-94=0


{ x=◦ ×=
-
✗ 3-94=0

✗ 1×2-91=0 X -


3=0 ✗ =3
✗ 1×-331×+31=0 .

}
Solve for the
Eg :
lengths of the sides of the right triangle below
e
✗+ "
✗ +2 Recall : Pythagorean theorem
b
c%a762
a

DO ,

1×1-412=1/21-1×+212
✗
✗ 21-8×1-16 = ✗21-1174×1-4=2×21-4×+4
0=2×21-4×+4 -

✗ 2-8×1-16
◦
0=(11-611×1-2)
= ✗ 2-4×-12 →

{ ×

✗ +2=0
-6=0 11=6
✗ = -2
Reject
lengths
this since
can't be -

ve

=) The side
lengths are 10
g

6
-
Exercises :

1 Solve the equation
a. ✗ 2-4×+3=0 b. ✗ 2=511
C. 16×2-49=0 d- 2×2-8×-3=0 e. ✗ 3=10×4-3911
2 If -1 is one solution of ✗ 21-4×1-5=0

find the other one .