boomerang285
sqrt x+10 = x-2
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boomerang285
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\[\sqrt{x+10}=x-2\]
Nameless
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you first need to get rid of the radical.
Nameless
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so square both sides.
boomerang285
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ok
Nameless
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so you should have x+10 = x^2-4x-4 ?
Nameless
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now move your terms all to one side and set it equal to 0 and factor or use the quadratic to solve for x
boomerang285
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ok
Nameless
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actualy... i typoed.. it should be \[x+10=x^{2}-4x+4\]
boomerang285
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ok i was confused trying to work that out
Nameless
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and the only reason i move to the right is cause I dont want to multiply by a negative 1.
Nameless
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\[x ^{2}-5x-6=0\]
Nameless
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you know how to factor or do the quadratic?
boomerang285
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i have my text book here, these confuse me so much
Nameless
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ok. i would factor this. so you set up parens.. ( )( ) right? fill them with (x )(x ). now we are looking for 2 numbers that multiply to 6 and add to -5
Nameless
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so 2 and 3 seem to work nice.. but they are not gonna get us to where we want because of the negative 5. so we try 1 and 6. (x 1)(x 6) right?
boomerang285
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so 2 and 3
boomerang285
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ok
Nameless
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now since we know we a negative 5.. we know that 6 has to be negative. so (x+1)(x-6)
boomerang285
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so (x+1)(x-6)
Nameless
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solve those as separate equations .. x-6=0 and x+1=0
boomerang285
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x=6 x=-1
Nameless
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yep. now.. the most important.. you need to plus these numbers back into the equation.. to check if both answers work.
boomerang285
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ok
Nameless
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\[\sqrt{-1+10} =-1 - 2\]
Nameless
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\[\sqrt{6+10} = 6-2\]
boomerang285
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sqrt9=-3 sqrt16 = 4
Nameless
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so x = 6 is your only answer? does that make sense?
Nameless
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\[\sqrt{9} =3 \neq -1-2 = -3\] \[\sqrt{16}=4 = 6-2=4\]