Question 408261: hi i need help with quadratic equations and i need steps please
1)(x-6)(x-4)=0
2)(x+2)(x-2)=0
3)(x+6)(x+3)=0
4)(x-4)(x+7)+0
5)x(-7)=0
6)(x-5)(x-2)=0
7)(x-8)(x-6)=0
8)(x+4)(x-4)=0
9)x(x+5)=0
10)(x-1)(x-8=0
11)(x-2)(2x-1)=0
12)(3x-2)(x+3)=
13)(x+5)(4x-1)=0
14)(2x-1)(3x+1)=0
15)4x(2x-3)=0
i need steps to please
Answer by MathLover1(20849) (Show Source):
You can put this solution on YOUR website!
1)(x-6)(x-4)=0
the first one inside the brackets (x-6)=0 , and the second one inside the brackets (x-4)=0; the first one equals x=6, the second one equals x=4
2)(x+2)(x-2)=0
the first one inside the brackets (x+2)=0 , and the second one inside the brackets (x-2)=0; the first one equals x=-2, the second one equals x=2
3)(x+6)(x+3)=0
the first one inside the brackets (x+6)=0 , and the second one inside the brackets (x+3)=0; the first one equals x=-6, the second one equals x=-3
4)(x-4)(x+7)=0
the first one inside the brackets (x-4)=0 and the second one inside the brackets
(x+3)=0; the first one equals x=4, the second one equals x=-3
5)x(-7)=0
since (-7) is not equal to zero, then x must be zero; x=0
6)(x-5)(x-2)=0
the first one inside the brackets (x-5)=0 and the second one inside the brackets
(x-2)=0; the first one equals x=5, the second one equals x=2
7)(x-8)(x-6)=0
the first one inside the brackets (x-8)=0 and the second one inside the brackets
(x-6)=0; the first one equals x=8, the second one equals x=6
8)(x+4)(x-4)=0
the first one inside the brackets (x+4)=0 and the second one inside the brackets
(x-4)=0; the first one equals x=-4, the second one equals x=4
9)x(x+5)=0
the first one x=0, second one (x+5)=0; therefore second one x=-5
10)(x-1)(x-8=0
the first one inside the brackets (x-1)=0 and the second one inside the brackets
(x-8)=0; the first one equals x=1, the second one equals x=8
11)(x-2)(2x-1)=0
the first one inside the brackets (x-2)=0 and the second one inside the brackets
(2x-1)=0; the first one equals x=2, the second one equals 2x=1..therefore x=1/2
12)(3x-2)(x+3)=0
the first one inside the brackets (3x-2)=0 and the second one inside the brackets
(x+3)=0; the first one equals 3x=2...or x=2/3, the second one equals x=-3
13)(x+5)(4x-1)=0
the first one inside the brackets (x+5)=0 and the second one inside the brackets
(4x-1)=0; the first one equals x=-5, the second one equals 4x=1..therefore x=1/4
14)(2x-1)(3x+1)=0
the first one inside the brackets (2x-1)=0 and the second one inside the brackets
(3x+1)=0; the first one equals 2x=1 or x=1/2, the second one equals 3x=-1..therefore x=-1/3
15)4x(2x-3)=0
the first one 4x=0 only if x=0
the second one inside the brackets
(2x-3)=0;......2x=3 or x=3/2
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