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4.8 Quadratic formula and the discriminant
4.8 Warm up
4.8 Quadratic formula and the discriminant
If we solve the quadratic formula for x
ax  bx  c  0
2
We get an equation to find x, given any quadratic
equation!!
 b  b  4ac
x
2a
2
4.8 Quadratic formula and the discriminant
Solve x2 + 3x = 2.
x2 + 3x = 2
x2 + 3x – 2 = 0
x = – b + b2 – 4ac
2a
x = – 3 + 32 – 4(1)(–2)
2(1)
x = – 3 + 17
2
Write original equation.
Write in standard form.
Quadratic formula
Identify a, b, and c
a = 1, b = 3, c = –2
Simplify.
ANSWER
The solutions are x = – 3 + 17
2
x = – 3 – 17
– 3.56.
2
0.56 and
4.8 Quadratic formula and the discriminant
Solve 25x2 – 18x = 12x – 9.
25x2 – 18x = 12x – 9.
Write original equation.
Write in standard form.
25x2 – 30x + 9 = 0.
x = 30 + (–30)2– 4(25)(9) a = 25, b = –30, c = 9
2(25)
30 + 0
x = 50
x = 53
ANSWER
The solution is 3
5
Simplify.
Simplify.
4.8 Quadratic formula and the discriminant
Solve –x2 + 4x = 5.
–x2 + 4x = 5
–x2 + 4x – 5 = 0.
x = – 4+ 42– 4(– 1)(– 5)
2(– 1)
– 4+ – 4
x= –2
– 4+ 2i
x=
–2
x=2+i
ANSWER
The solution is 2 + i and 2 – i.
Write original equation.
Write in standard form.
a = –1, b = 4, c = –5
Simplify.
Rewrite using the
imaginary unit i.
Simplify.
4.8 Quadratic formula and the discriminant
Use the quadratic formula to solve the equation.
x2 = 6x – 4
x2 = 6x – 4
x2 – 6x + 4 = 0
x = – b + b2 – 4ac
2a
x = + 3 + 20
2
x = – (– 6) + (– 6)2 – 4(1)(4)
2(1)
20
The solutions are x = 3 +
x=
3–
20
2
2
=3–
5
=3+
5
and
4.8 Quadratic formula and the discriminant
Use the quadratic formula to solve the equation.
2.
4x2 – 10x = 2x – 9
x = – b + b2 – 4ac
2a
SOLUTION
4x2 – 10x = 2x – 9
4x2 – 12x + 9 = 0
x = – (– 12) + (– 12)2 – 4(4)(9)
2(4)
x = 12 + 0
8
The solution is 3 = 1 1 .
2
2
4.8 Quadratic formula and the discriminant
Use the quadratic formula to solve the equation.
3.
7x – 5x2 – 4 = 2x + 3
SOLUTION
7x – 5x2 – 4 = 2x + 3
– 5x2 + 5x – 7 = 0
x = – (5) +
(5)2 – 4(– 5)(–7)
2(– 5)
x = – b + b2 – 4ac
2a
x = – 5 + –115
– 10
x = 5 + i 115
10
ANSWER
The solutions are
5 + i 115 and
10
5 – i 115 .
10
4.8 Quadratic formula and the discriminant
 b  b  4ac
x
2a
2
Lets look at just
a part of the
quadratic
formula
What effect does this
part have on your
solution if it is positive?
If it is negative?
If it is zero?
4.8 Quadratic formula and the discriminant
•disc. < 0
two imaginary solutions
•disc. > 0
two real solutions
•disc. = 0
one real solution
Find the discriminant and type of solutions for
–x2 + 4x = 5.
a = –1, b = 4, c = –5
b 2  4ac  42  4(1)(5)
= 16 – 20
= – 4,
two imaginary solutions
4.8 Quadratic formula and the discriminant
Find the discriminant of the quadratic equation and
give the number and type of solutions of the
equation.
a. x2 – 8x + 17 = 0
b. x2 – 8x + 16 = 0
c. x2 – 8x + 15 = 0
Discriminant
Solution(s)
SOLUTION
Equation
ax2 + bx + c = 0
b2 – 4ac
x = – b+ b2– 4ac
2a
a. x2 – 8x + 17 = 0
(–8)2 – 4(1)(17) = – 4 Two imaginary: 4 + i
b. x2 – 8x + 16 = 0
(–8)2 – 4(1)(16) = 0
One real: 4
c. x2 – 8x + 15 = 0
(–8)2 – 4(1)(15) = 4
Two real: 3,5
4.8 Quadratic formula and the discriminant
Find the discriminant of the quadratic equation and give
the number and type of solutions of the equation.
2x2 + 4x – 4 = 0
SOLUTION
Equation
Discriminant
ax2 + bx + c = 0
b2 – 4ac
2x2 + 4x – 4 = 0
42 – 4(2)(– 4 )
= 48
Solution(s)
2
x = – b+ b – 4ac
2a
Two real solutions
4.8 Quadratic formula and the discriminant
3x2 + 12x + 12 = 0
SOLUTION
Equation
ax2 + bx + c = 0
3x2 + 12x + 12 = 0
Discriminant
b2 – 4ac
122 – 4(12)(3 )
=0
Solution(s)
2
x = – b+ b – 4ac
2a
One real solution
4.8 Quadratic formula and the discriminant
8x2 = 9x – 11
SOLUTION
Equation
ax2 + bx + c = 0
8x2 – 9x + 11 = 0
Discriminant
b2 – 4ac
(– 9)2 – 4(8)(11 )
= – 271
Solution(s)
2
x = – b+ b – 4ac
2a
Two imaginary
solutions
4.8 Quadratic formula and the discriminant
7x2 – 2x = 5
SOLUTION
Equation
ax2 + bx + c = 0
7x2 – 2x – 5 = 0
Discriminant
b2 – 4ac
(– 2)2 – 4(7)(– 5 )
= 144
Solution(s)
2
x = – b+ b – 4ac
2a
Two real solutions
4.8 Quadratic formula and the discriminant
4x2 + 3x + 12 = 3 – 3x
SOLUTION
Equation
ax2 + bx + c = 0
4x2 + 6x + 9 = 0
Discriminant
b2 – 4ac
(6)2 – 4(4)(9)
= – 108
Solution(s)
2
x = – b+ b – 4ac
2a
Two imaginary
solutions
4.8 Quadratic formula and the discriminant
3x – 5x2 + 1 = 6 – 7x
SOLUTION
Equation
Discriminant
b2 – 4ac
ax2 + bx + c = 0
– 5x2 + 4x– 5 = 0
(4)2 – 4(– 5)(– 5)
=0
HOMEWORK 4.8
p. 296 #3-60 EOP
Solution(s)
2
x = – b+ b – 4ac
2a
One real solution