Download 7.3 Notes - West Ada

7.4 Notes
SOLVING QUADRATICS
Solving Quadratics

Finding x-intercepts
 Finding
zero

the value of x to make the equation equal
Degree is 2
 Two
values of x
Solving by Factoring

Factor the quadratic

Set each factor equal to zero
 Solve

for x
3x2 – 24x
 3x(x-8)
 3x
=0
x
=0
x–8=0
x=8
Difference of Squares (b = 0)

4x2 – 25
 (2x
 2x
+ 5)(2x – 5)
+5=0
x
 2x
= - 5/2
–5=0
x
= 5/2
 x2
+ 3x – 40
 (x
+ 8)(x – 5) = 0
 (x
+ 8) = 0
x
 (x
=-8
– 5) = 0
x
=5
Factoring when a is NOT 1

2x2 + 7x + 6 = 0

2x2
+ 3x + 4x + 6 = 0
 x(2x
 (2x

+ 3) + 2(2x + 3) = 0
+ 3)(x + 2) = 0
 2x
x
+3=0
= - 3/2
2*6 = 12
7
2*6 = 12
2+6=8
3*4 = 12
3+4=7
x+2=0
x = -2
-4x2
– 2x + 56 = 0

-2(2x2 + x – 28) = 0

-2(2x2 + 8x – 7x – 28) = 0

-2(2x(x + 4) – 7(x + 4) = 0

-2(x + 4)(2x – 7) = 0

x+4=0

x = -4
2x – 7 = 0
x = 7/2
2*-28 = -56 1
8*-7 = -56
8 + (-7) = 1
Solving with the Quadratic Formula

ax2 + bx + c = 0
−𝑏± 𝑏 2 −4𝑎𝑐
2𝑎

Formula: 𝑥 =

3x2 – 2x – 7 = 0

a=3

𝑥=
𝑥

b = -2 c = -7
2± (−2)2 −4∗3∗−7
2∗3
=
2± 4+84
6
2±3 10
6
=
2± 90
6