x^2+2x-3=0
(x+3)(x-1) = 0
x=-3 or x=1
on a number line circle the -3 and the 1
Solve by Graphing
x^2+2x-3=0
9 questions
quadratic systems quiz part 1
unit 5 lesson 8
please post all answers
5 answers
I have just finished and here are the answers
1:c (1, -3)
2:b (6.2, 9.1)
3:a (1-7i)
4:d (no)
5:a (0,3)
6:a (-3/4i,3/4i)
7:c (-4,0 -1,3)
8:a (tiny grey)
9:b (5i)
1:c (1, -3)
2:b (6.2, 9.1)
3:a (1-7i)
4:d (no)
5:a (0,3)
6:a (-3/4i,3/4i)
7:c (-4,0 -1,3)
8:a (tiny grey)
9:b (5i)
Step by step solution
Factoring this quadratic equation.
_____________
Remark:
- x + 3 x = 2 x
_____________
x² + 2 x - 3 = x² - x + 3 x - 3 =
( x² - x ) + ( 3 x - 3 )=
x ( x - 1 ) + 3 ( x - 1) =
( x - 1 ) ( x + 3 )
x² + 2 x - 3 = 0
is same as
( x - 1 ) ( x + 3 ) = 0
The equation will be equal to zero when the expressions in parentheses are equal to zero.
First solution.
x - 1 = 0
Add 1 to both sides.
x = 1
Second solution.
x + 3 = 0
Subtract 3 to both sides.
x = - 3
The solutions are:
x = - 3 and x = 1
Factoring this quadratic equation.
_____________
Remark:
- x + 3 x = 2 x
_____________
x² + 2 x - 3 = x² - x + 3 x - 3 =
( x² - x ) + ( 3 x - 3 )=
x ( x - 1 ) + 3 ( x - 1) =
( x - 1 ) ( x + 3 )
x² + 2 x - 3 = 0
is same as
( x - 1 ) ( x + 3 ) = 0
The equation will be equal to zero when the expressions in parentheses are equal to zero.
First solution.
x - 1 = 0
Add 1 to both sides.
x = 1
Second solution.
x + 3 = 0
Subtract 3 to both sides.
x = - 3
The solutions are:
x = - 3 and x = 1
Answrr is c
( 1, - 3 )
( 1, - 3 )
bramboy is correct
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