1. write 15x 2 + 6x = 14x 2 – 12 in standard form. 2. evaluate b 2 – 4ac when a = 3, b = –6,...
TRANSCRIPT
1. Write 15x2 + 6x = 14x2 – 12 in standard form.
2. Evaluate b2 – 4ac when a = 3, b = –6, and c = 5.
ANSWER
ANSWER –24
x2 + 6x +12 = 0
Classwork Answers…4.8 (13-27 odd)
4.6 (multiples of 3)
a
acbbx
2
42
When solving quadratic equations, we’re looking for the x values where the graph crosses the x axis
One of the methods we use to solve quadratic equations is called the Quadratic Formula
a
acbbx
2
42
Using the a, b, and c from ax2 + bx + c = 0
Must be equal to zero
EXAMPLE 1 Solve an equation with two real solutions
Solve x2 + 3x = 2.x2 + 3x = 2 Write original equation.
x2 + 3x – 2 = 0 Write in standard form.
x =– b + b2 – 4ac2a
Quadratic formula
x =– 3 + 32 – 4(1)(–2) 2(1)
a = 1, b = 3, c = –2
Simplify.x = – 3 + 172
56.02
173
x or 56.3
2
173
x
EXAMPLE 2 Solve an equation with one real solutions
Solve 25x2 – 18x = 12x – 9.25x2 – 18x = 12x – 9. Write original equation.
Write in standard form.
x =30 + (–30)2– 4(25)(9)2(25)
a = 25, b = –30, c = 9
Simplify.
25x2 – 30x + 9 = 0.
x =30 + 0
50
x = 35 Simplify.
35The solution is
ANSWER
EXAMPLE 3 Solve an equation with imaginary solutions
Solve –x2 + 4x = 5.–x2 + 4x = 5 Write original equation.
Write in standard form.
x =–4 + 42 – 4(–1)(–5)2(–1)
a = –1, b = 4, c = –5
Simplify.
–x2 + 4x – 5 = 0.
x =–4 + –4
–2–4 + 2i
x = –2
Simplify.
Rewrite using the imaginary unit i.
x = 2 + i
The solution is 2 + i and 2 – i.
ANSWER
GUIDED PRACTICE for Examples 1, 2, and 3
Use the quadratic formula to solve the equation.
x2 = 6x – 41.
x2 – 6x + 4 = 0
a = 1 b = -6 c = 4
x =– b + b2 – 4ac2a
)1(2
)4)(1(4)6(6 2 x
2
206 x
2
526 x
53x
2
)53(2 x
GUIDED PRACTICE for Examples 1, 2, and 3
Use the quadratic formula to solve the equation.
4x2 – 10x = 2x – 92.
4x2 – 12x + 9 = 0
a = 4 b = -12 c = 9
x =– b + b2 – 4ac2a
)4(2
)9)(4(4)12(12 2 x
8
012 x
8
12x
2
3x
EXAMPLE 4 Use the discriminant
EquationDiscriminant
Solution(s)
If the quadratic equation is in the standard form ax2 + bx + c = 0
b2 – 4ac
a. x2 – 8x + 17 = 0 ( –8)2 – 4(1)(17) = –4 Two imaginary
b. x2 – 8x + 16 = 0 (–8)2 – 4(1)(16) = 0 One real
b. x2 – 8x + 15 = 0 (–8)2 – 4(1)(15) = 4 Two real
The discriminant can be found using b2 – 4ac
If b2 – 4ac < 0 There are Two Imaginary solutions
If b2 – 4ac = 0 There is One Real solution
If b2 – 4ac > 0 There are Two Real solutions
GUIDED PRACTICE for Example 4
Find the discriminant of the quadratic equation and give the number and type of solutions of the equation.
4. 2x2 + 4x – 4 = 0 5. 3x2 + 12x + 12 = 0
6. 8x2 = 9x – 11 7. 4x2 + 3x + 12 = 3 – 3x
(4)2 – 4(2)(-4) = 48
b2 – 4ac
Two Real solutions
(12)2 – 4(3)(12) = 0
b2 – 4ac
One Real solution
(-9)2 – 4(8)(11) = -271
b2 – 4ac
Two Imaginary solutions
(6)2 – 4(4)(9) = -108
b2 – 4ac
Two Imaginary solutions
8x2 – 9x + 11 = 0 4x2 + 6x + 9 = 0
Classwork Assignment:WS 4.8 (1-27 odd)