Download Notes on Solving Quadratic Equations by Factoring

Solving Quadratic
Equations by Factoring
Quadratic Equations
• Can be written in the form ax2 + bx + c = 0.
• a, b and c are real numbers and a  0.
• This is referred to as standard form.
Zero Factor Theorem
• If a and b are real numbers and ab = 0, then a =
0 or b = 0.
• This theorem is very useful in solving quadratic
equations.
Steps for Solving a Quadratic Equation by
Factoring
1)
2)
3)
4)
5)
Write the equation in standard form.
Factor the quadratic completely.
Set each factor containing a variable equal to 0.
Solve the resulting equations.
Check each solution in the original equation.
Example
Solve x2 – 5x = 24.
• First write the quadratic equation in standard form.
x2 – 5x – 24 = 0
• Now we factor the quadratic using techniques from
the previous sections.
x2 – 5x – 24 = (x – 8)(x + 3) = 0
• We set each factor equal to 0.
x – 8 = 0 or x + 3 = 0, which will simplify to
x = 8 or x = – 3
Continued.
Example Continued
• Check both possible answers in the original
equation.
82 – 5(8) = 64 – 40 = 24
true
(–3)2 – 5(–3) = 9 – (–15) = 24
true
• So our solutions for x are 8 or –3.
Example
Solve 4x(8x + 9) = 5
• First write the quadratic equation in standard
form.
32x2 + 36x = 5
32x2 + 36x – 5 = 0
• Now we factor the quadratic.
32x2 + 36x – 5 = (8x – 1)(4x + 5) = 0
• We set each factor equal to 0.
8x – 1 = 0 or 4x + 5 = 0
1
5

8x = 1 or 4x = – 5, which simplifies to x = 8 or 4 .
Continued.
Example Continued
• Check both possible answers in the original equation.
      
1
4
8
1
1
8
9  4
8
8

1
1
1  9   4 8 (10)  2 (10)  5
      
4 
true
 
5
5
5
5
8   9  4   10  9   4  (1)  (5)(1)  5
4
4
4
4
true
• So our solutions for x are
1
8
or
5

4
.
Recall earlier we found the x-intercept(s) of
linear equations by letting y = 0 and solving
for x.
The same method works for x-intercepts in
quadratic equations.
Note: When the quadratic equation is written
in standard form, the graph is a parabola
opening up (when a > 0) or down (when a <
0), where a is the coefficient of the x2 term.
The intercepts will be where the parabola
crosses the x-axis.
Example
Find the x-intercepts of the graph of f(x)= 4x2
+ 11x + 6.
The equation is already written in standard form, so
we let f(x) = 0, then factor the quadratic in x.
0 = 4x2 + 11x + 6 = (4x + 3)(x + 2)
We set each factor equal to 0 and solve for x.
4x + 3 = 0 or x + 2 = 0
4x = –3 or x = –2
x = –¾ or x = –2
So the x-intercepts are the points (–¾, 0) and (–2, 0).
Quadratic Equations
and Problem Solving
General Strategy for Problem Solving
1) Understand the problem
• Read and reread the problem
• Choose a variable to represent the
unknown
• Construct a drawing, whenever possible
• Propose a solution and check
2) Translate the problem into an equation
3) Solve the equation
4) Interpret the result
• Check proposed solution in problem
• State your conclusion
Finding an Unknown Number
Example
The product of two consecutive positive integers is 132. Find the
two integers.
1.) Understand
Read and reread the problem. If we let
x = one of the unknown positive integers, then
x + 1 = the next consecutive positive integer.
Continued
Finding an Unknown Number
Example continued
2.) Translate
The product of
is
132
=
132
two consecutive positive integers
x
•
(x + 1)
Continued
Finding an Unknown Number
Example continued
3.) Solve
x(x + 1) = 132
x2 + x = 132
x2 + x – 132 = 0
(x + 12)(x – 11) = 0
x + 12 = 0 or x – 11 = 0
x = –12 or x = 11
(Distributive property)
(Write quadratic in standard form)
(Factor quadratic polynomial)
(Set factors equal to 0)
(Solve each factor for x)
Continued
Finding an Unknown Number
Example continued
4.) Interpret
Check: Remember that x is suppose to represent a positive
integer. So, although x = -12 satisfies our equation, it cannot be a
solution for the problem we were presented.
If we let x = 11, then x + 1 = 12. The product of the two numbers
is 11 · 12 = 132, our desired result.
State: The two positive integers are 11 and 12.
The Pythagorean Theorem
Pythagorean Theorem
In a right triangle, the sum of the squares
of the lengths of the two legs is equal to
the square of the length of the
hypotenuse.
(leg a)2 + (leg b)2 = (hypotenuse)2
leg a
hypotenuse
leg b
The Pythagorean Theorem
Example
Find the length of the shorter leg of a right triangle if the longer leg
is 10 miles more than the shorter leg and the hypotenuse is 10 miles
less than twice the shorter leg.
1.) Understand
Read and reread the problem. If we let
x = the length of the shorter leg, then
x + 10 = the length of the longer leg and
2x – 10 = the length of the hypotenuse.
2 x - 10
x
x + 10
Continued
The Pythagorean Theorem
Example continued
2.) Translate
By the Pythagorean Theorem,
(leg a)2 + (leg b)2 = (hypotenuse)2
x2 + (x + 10)2 = (2x – 10)2
3.) Solve
x2 + (x + 10)2 = (2x – 10)2
x2 + x2 + 20x + 100 = 4x2 – 40x + 100
2x2 + 20x + 100 = 4x2 – 40x + 100
(multiply the binomials)
(simplify left side)
0 = 2x2 – 60x
(subtract 2x2 + 20x + 100 from both sides)
0 = 2x(x – 30)
(factor right side)
x = 0 or x = 30 (set each factor = 0 and solve)Continued
The Pythagorean Theorem
Example continued
4.) Interpret
Check: Remember that x is suppose to represent the length of
the shorter side. So, although x = 0 satisfies our equation, it
cannot be a solution for the problem we were presented.
If we let x = 30, then x + 10 = 40 and 2x – 10 = 50. Since 302 +
402 = 900 + 1600 = 2500 = 502, the Pythagorean Theorem
checks out.
State: The length of the shorter leg is 30 miles. (Remember that
is all we were asked for in this problem.)