SOLUTION: The question says solve:
x(x+1)= -2(x-4)=2
This is what I did, can you tell me if I am on the right track, please?
x(x+1)= -2(x-4)=2
x*2+1x= -2x+8+2
x*2+x+2x-8-2=0
x*2+3-10=0
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Distributive-associative-commutative-properties
-> SOLUTION: The question says solve:
x(x+1)= -2(x-4)=2
This is what I did, can you tell me if I am on the right track, please?
x(x+1)= -2(x-4)=2
x*2+1x= -2x+8+2
x*2+x+2x-8-2=0
x*2+3-10=0
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Question 73624: The question says solve:
x(x+1)= -2(x-4)=2
This is what I did, can you tell me if I am on the right track, please?
x(x+1)= -2(x-4)=2
x*2+1x= -2x+8+2
x*2+x+2x-8-2=0
x*2+3-10=0
(x+5) (x-2)
x+5=0 x-2=0
x=-5 and x=-2 Found 2 solutions by Earlsdon, bucky:Answer by Earlsdon(6294) (Show Source):
You can put this solution on YOUR website! Well, there are a couple of discrepancies I see here.
1) Although your first equation is written as:
x(x+1)=-2(x-4)=2, I assume that there is a typo and that you meant to write: Right?
The second thing that I notice is that you have not handled the (+ & -) signs correctly. Let's step through the process: Apply the distributive property to both sides. Combine like-terms. Subtract 10 from both sides. Factor this quadratic equation. Apply the zero product principle. and/or
Solutions:
x = 2 or x = -5
You can put this solution on YOUR website! The question says solve: <---typing error - the second = sign is supposed to be "+"
.
This is what I did, can you tell me if I am on the right track, please?
. <--- same comment as above. you correct this OK in the next step <--- no need to show the 1 as the multiplier of x. <--- OK <--- should be +3x, not just 3 <---- should show this product as equal to 0 <---- OK and <--- the x = -5 is correct but you have a very minor mistake in
saying x = -2. Check your work on solving x-2 = 0.
.
Good job! You seem to understand all the steps you have to do to solve this equation, and
you also understand factoring quadratic equations.