Download Solving Quadratic Equations by Finding Square Roots

9.1 SOLVING QUADRATIC
EQUATIONS BY FINDING SQUARE
ROOTS
 Students will first practice how to evaluate square roots
and then will solve a quadratic equation by finding the
square root.
WHAT IS A SQUARE ROOT?
If a number square (b2) = another number (a),
then b is the square root of a.
Example: If 32 = 9, then 3 is the square root
of 9
SOME BASICS…
 All positive numbers have two square roots:
1. The 1st is a positive square root, or principal square root.
2. The 2nd is a negative square root (see Interactive Notebook)
 Square roots are written with a radical symbol
 You can show both square roots by using the “plusminus” symbol ±
RADICAL
3
Radical sign
Radicand
(See Interactive Notebook)
FIND THE SQUARE ROOT OF NUMBERS
64  8
 64  8
0 0
 0.25  .5
4
(undefined )
PERFECT SQUARES: NUMBERS WHOSE SQUARE ROOTS
ARE INTEGERS OR QUOTIENTS OF INTEGERS.
1 1
4 2
9 3
16  4
25  5
36  6
49  7
64  8
81  9
100  10
121  11
144  12
169  13
EVALUATE A RADICAL EXPRESSION
The square root symbol is a grouping symbol. Operations
inside the radical symbol must be performed before the
square root is evaluated
Evaluate b 2  4ac when a  1, b  2, and c  3.
b 2  4ac  (2) 2  4(1)( 3)  4  4(3)
 4  12  16  4
QUADRATIC EQUATIONS
Standard form: ax2 + bx + c = 0
a is the leading coefficient and cannot be equal to
zero.
If the value of b were equal to zero, the equation
becomes ax2 + c = 0.
We can solve equations is this form by taking the
square root of both sides.
KEY CONCEPTS
When x2 = d
If d > 0, then x2 = d has two solutions
If d = 0, then x2 = d has one solution
If d < 0, then x2 = d has no real solution
SOLVING QUADRATICS
Solve each equation.
a. x2=4 b. x2=5 c. x2=0 d. x2=-1
x2=4 has two solutions, x = 2, x = -2
x2=5 has two solutions, x =√5, x =- √5
x2=0 has one solution, x = 0
x2=-1 has no real solution
SOLVE BY REWRITING EQUATION
Example: Solve 3x2 – 48 = 0
1. 3x2 – 48 + 48 = 0 + 48
2. 3x2 = 48
3. 3x2 / 3 = 48 / 3
4. x2 = 16
5. After taking square root of both sides,
6. x = ± 4
EXAMPLE
a. 2x2 = 8
Solve the equation.
2x2 = 8
SOLUTION
a. 2x2 = 8
x2 = 4
x = ± 4 = ± 2
ANSWER
Write original equation.
Divide each side by 2.
Take square roots of each side.
Simplify.
The solutions are – 2 and 2.
b. m2 – 18 = – 18
m2 = 0
m= 0
ANSWER
The solution is 0.
Write original equation.
Add 18 to each side..
The square root of 0 is 0.
c. b2 + 12 = 5
b2 = – 7
Write original equation.
Subtract 12 from each side.
ANSWER
Negative real numbers do not have real square roots.
So, there is no solution.
EXAMPLE
c2 – 25 = 0
Solve the equation.
SOLUTION
c2 – 25 = 0
c = ± 25 = ± 5
ANSWER
The solutions are – 5 and 5.
Write original equation.
Take square roots of each side.
Simplify.
EXAMPLE
Solve 4z2 = 9.
SOLUTION
4z2 = 9.
Write original equation.
9
z2 = 4
z = ±
Divide each side by 4.
9
4
z=± 3
2
ANSWER
The solutions are – 3 and 3
2
2
Take square roots of each side.
Simplify.
EXAMPLE
An engineering student is in an “egg dropping
contest.” The goal is to create a container for
an egg so it can be dropped from a height of 32
feet without breaking the egg. To the nearest
tenth of a second, about how long will it take
for the egg’s container to hit the ground?
Assume there is no air resistance.
h  16t  s
2
The question asks to find the time it takes for the
container to hit the ground.
Initial height (s) = 32 feet
Height when its ground (h) = 0 feet
Time it takes to hit ground (t) = unknown
h  16t  s
2
1. Substitute
0 = -16t2 + 32
2. -32 + 0 = -16t2 + 32 – 32
3. -32 = -16t2
4. -32 / -16 = -16t2 / -16
5. 2 = t2
t = √2 seconds or approx. 1.4 seconds
ASSIGNMENT
9.1 w/s