Lesson 9-4 Warm-Up
ALGEBRA 1
“Solving Quadratic Equations”
(9-4)
What is a
quadratic
equation?
quadratic equation: a quadratic function (y = ax2 + bx + c) where y = 0 and
written in the standard form:
ax2 + bx + c = 0
Tip: Recall that to find the x-intercepts of a quadratic function, find x when y = 0.
Therefore, the solution(s) of a quadratic equation are also the x-intercepts
of the graph of the quadratic function (which is a parabola).
What are the
“roots of the
equation” or the
“zeros of the
function”?
Since a parabola can have two, one, or no points that cross the x-axis (xintercepts), a quadratic equation can have two, one, or no solutions. The
solutions of a quadratic equation and the related x-intercepts are called the
roots of the equation or the zeros of the function.
ALGEBRA 1
“Solving Quadratic Equations”
(9-4)
To solve a quadratic equation, substitute 0 for y and solve for x.
How do you
solve a quadratic Example: y = x2 - 4
equation when b
x2 - 4 = 0
Make y = 0 and write it in standard form (zero on
= 0?
right side of equal sign)
+ 4 4
x2
=4
x2 = 4
x = 2
Add 4 to both sides
Simplify
Find the square root of each side to eliminate the
square (square root of a square is the number –
the operations cancel each other out)
The square root of a number can be negative (-•=+)
The solutions, or x-intercepts,
of y = x2 - 4 are 2 and -2.
ALGEBRA 1
Solving Quadratic Equations
LESSON 9-4
Additional Examples
Solve each equation by graphing the related function.
a. 2x2 = 0
Graph y = 2x2
There is one
solution, x = 0.
b. 2x2 + 2 = 0
Graph y = 2x2 + 2
There is no solution.
c. 2x2 – 2 = 0
Graph y = 2x2 – 2
There are two
solutions, x = ±1.
ALGEBRA 1
Solving Quadratic Equations
LESSON 9-4
Additional Examples
Solve 3x2 – 75 = 0.
3x2 – 75 + 75 = 0 + 75
Add 75 to each side.
3x2 = 75
x2 = 25
x2 =
x=±5
Divide each side by 3.
25
Square root both sides to isolate the x.
Simplify.
ALGEBRA 1
“Solving Quadratic Equations”
(9-4)
How can you
solve a real
world problem
involving square
roots?
To solve a real-world problem or equation involving square roots, negative
solutions are not reasonable and thus should be eliminated (not used).
ALGEBRA 1
Solving Quadratic Equations
LESSON 9-4
Additional Examples
A museum is planning an exhibit that will contain a large
globe. The surface area of the globe will be 315 ft2. Find the radius
of the sphere producing this surface area. Use the equation
S = 4 r2, where S is the surface area and r is the radius.
S = 4 r2
315 = 4 r 2
4
4
Substitute 315 for S.
315
= r2
4
Divide each side by p (4).
315
=r
(4)
Square root each side to isolate the r.
 5.00668588
r
Use a calculator. Since a negative solution isn’t
reasonable (doesn’t make sense), only the positive
solution should be used.
The radius of the sphere is about 5 ft.
ALGEBRA 1
Solving Quadratic Equations
LESSON 9-4
Lesson Quiz
1. Solve each equation by graphing the related function. If the equation
has no solution, write no solution.
a. 2x2 – 8 = 0
±2
b. x2 + 2 = –2 no solution
2. Solve each equation by finding square roots.
a. m2 – 25 = 0 ±5
b. 49q2 = 9
3
±7
3. Find the speed of a 4-kg bowling ball with a kinetic energy of 160
joules. Use the equation E = 1 ms2, where m is the object’s mass in kg,
2
E is its kinetic energy, and s is the speed in meters per second.
about 8.94 m/s
ALGEBRA 1