Using the Zero-Product Property to Solve a Quadratic
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Transcript Using the Zero-Product Property to Solve a Quadratic
Using the Zero-Product
Property to Solve a Quadratic
x–Intercepts, Solutions, Roots, and
Zeros in Quadratics
x-intercept(s): Where the graph of y=ax2+bx+c
crosses the x-axis. The value(s) for x that makes
a quadratic equal 0.
Solution(s) OR Roots: The value(s) of x that
satisfies 0=ax2+bx+c.
Zeros: The value(s) of x that make ax2+bx+c equal
0.
Zero Product Property
If a . b = 0, then a and or b is equal to 0
Ex: Solve the following equation below.
0 = ( x + 14 )( 6x + 1 )
x 14 0
x 14
6x 1 0
6 x 1
OR
x 16
Would you rather solve the equation above or this:
0 = x2 + 25x + 14 ?
Solving a Quadratic: Factoring
Solve: 0 2 x2 5x 12
Factor to
rewrite as
a product
c
4
8x
Product
(2x2)(-12)
-12
-24x2
ax2c
GCF ___
x
Use the ZeroProduct
Property
2x2
-3x
ax2
2x
-3
0 x 4 2x 3
x 4 0 2x 3 0
x 4
2x 3
3
x 2
8x
bx
-3x
5x
Sum
x 4 or
3
2
Solving a Quadratic: Already Factored
Solve: 6x x 7 0
The equation is already factored AND it equals 0. Half
the work is already done. Just use the Zero-Product
Property
6x x 7 0
x7 0
6x 0
x0
x7
x 0 or 7
Solving a Quadratic: Making Sure to
Isolate 0
Solve: 35 12 x 2 44 x
Factor to rewrite as a product
Product
c
5
Solve for 0 first!
35 12 x 2 44 x
0 12 x 2 44 x 35
30x
(2x2)(-12)
35
420x2
2
12x
GCF 2x
___
2
ax2c
14x
30x
ax
Use the Zero-Product Property
6x
0 2x 5 6x 7
2x 5 0 6x 7 0
2 x 5
6 x 7
x 52
x 76
7
bx
44x
Sum
x or
5
2
7
6
14x
Solving a Quadratic: Make Sure to
Isolate 0
Solve: x 5 x 2 6
Factor to rewrite as a product
Product
c
Solve for 0 first!
x 5 x 2 6
-4
-4x
-4
(x2)(-4)
-4x2
Distribute
x2 3x 10 6 GCF ___
x
x 2 3x 4 0
Use the Zero-Product Property
x2
ax2c
x
ax2
x
0 x 4 x 1
x 1 0
x40
x 1
x4
1
-4x
bx
-3x
Sum
x 4 or 1
1x
Do we Need Another Method?
Use the Zero Product Property to find the roots of:
x 2 3x 7 0
Product
c
-7
(x2)(-7)
But this
parabola has
two zeros.
-7x2
ax2c
___
x2
ax2
IMPOSSIBLE
bx
-3x
Sum
Just because a quadratic is not factorable, does not mean it does
not have roots. Thus, there is a need for a new algebraic method
to find these roots.