I'm soo brain dead right now and reallly need help figuring out how to solve these problems. I'm not looking for answers, just the steps so I can understand what to do! ANy help at all would be so appreciated. (:
Solve by factoring.
1.) 3x²=x
2.) x²+14x + 49=0
3.) 14x²-29x-15=0
Solve by any method
4.) 4x²-9=0
5.) x²+4x=10x-9
6.) (x+5)(x-2)<0
Thanks SO much!
Solve by factoring.
1.) 3x²=x
2.) x²+14x + 49=0
3.) 14x²-29x-15=0
Solve by any method
4.) 4x²-9=0
5.) x²+4x=10x-9
6.) (x+5)(x-2)<0
Thanks SO much!
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3x^2 = x
3x^2 - x = 0
x(3x - 1) = 0
Therefore x = 0 or (3x - 1) = 0, i.e. x = 1/3
Answers: 0, 1/3
x^2 + 14x + 49 = (x+7)^2 = 0
Therefore x = -7
14x^2 - 29x - 15 = (7x+3)(2x-5) = 0
Therefore
7x+3 = 0
x = -3/7
or
2x-5 = 0
x = 5/2
4x^2 - 9 = 0
use the difference of two squares
(2x+3)(2x-3) = 0
Therefore x = 3/2 or -3/2
x^2 + 4x = 10x - 9
x^2 - 6x + 9 = (x-3)^2 = 0
Therefore x = 3
(x+5)(x-2) < 0
the function = 0 at x = -5 and x = 2
the function is less than zero between -5 and 2
Answer: -5 < x < 2
3x^2 - x = 0
x(3x - 1) = 0
Therefore x = 0 or (3x - 1) = 0, i.e. x = 1/3
Answers: 0, 1/3
x^2 + 14x + 49 = (x+7)^2 = 0
Therefore x = -7
14x^2 - 29x - 15 = (7x+3)(2x-5) = 0
Therefore
7x+3 = 0
x = -3/7
or
2x-5 = 0
x = 5/2
4x^2 - 9 = 0
use the difference of two squares
(2x+3)(2x-3) = 0
Therefore x = 3/2 or -3/2
x^2 + 4x = 10x - 9
x^2 - 6x + 9 = (x-3)^2 = 0
Therefore x = 3
(x+5)(x-2) < 0
the function = 0 at x = -5 and x = 2
the function is less than zero between -5 and 2
Answer: -5 < x < 2
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1. 3x^2 - x = 0
x(3x - 1) = 0
x=0 3x-1=0
so x= 0 and x=1/3
2. (x + 7)(x + 7) = 0
x = -7
4. (2x - 3)(2x + 3) = 0
x= -2/3 and x= 2/3
5. x^2 - 10x + 9 =0
(x -9)(x - 1) =0
x = 1 and x = 9
x(3x - 1) = 0
x=0 3x-1=0
so x= 0 and x=1/3
2. (x + 7)(x + 7) = 0
x = -7
4. (2x - 3)(2x + 3) = 0
x= -2/3 and x= 2/3
5. x^2 - 10x + 9 =0
(x -9)(x - 1) =0
x = 1 and x = 9