how to do these??
a) 3(x-10)^2 / 4 = 12
b) x-2 = 12 / x-2
c)5-2x / 4 = -2 1/2
d) (3x + 2)^2 -1 = 8
THANKYOU ! thankyou thankyou !!! I never got a chance to ask my math teacher about these ones, and I have a big test coming up tomorrow. Your help would be so fantastic!!
10 points to best answer!
a) 3(x-10)^2 / 4 = 12
b) x-2 = 12 / x-2
c)5-2x / 4 = -2 1/2
d) (3x + 2)^2 -1 = 8
THANKYOU ! thankyou thankyou !!! I never got a chance to ask my math teacher about these ones, and I have a big test coming up tomorrow. Your help would be so fantastic!!
10 points to best answer!
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a) 3(x-10)^2 / 4 = 12
>> 3(x-10)^2 = 4 x 12
>> (x-10)^2 = 48/3
>> (x-10)^2 = 16
>> x-10 = +- 4 [Square root]
If x-10 = 4, then x=14 OR if x-10= -4 , then x = 6 <<<
b) x-2 = 12 / x-2
>> (x-2)(x-2) =12
>> (x-2)^2 = 12
>> x-2 = +-√12
If x-2 =√12 , then x=2+√12 OR if x-2= -√12 , then x = 2-√12 <<<
d) (3x + 2)^2 -1 = 8
>> (3x + 2)^2 = 9
>> (3x + 2) = +-3
If 3x+2 = 3
>> 3x=1
>> x = 1/3
OR if 3x+2 = -3
>> 3x = -5
>> x = -5/3
>> 3(x-10)^2 = 4 x 12
>> (x-10)^2 = 48/3
>> (x-10)^2 = 16
>> x-10 = +- 4 [Square root]
If x-10 = 4, then x=14 OR if x-10= -4 , then x = 6 <<<
b) x-2 = 12 / x-2
>> (x-2)(x-2) =12
>> (x-2)^2 = 12
>> x-2 = +-√12
If x-2 =√12 , then x=2+√12 OR if x-2= -√12 , then x = 2-√12 <<<
d) (3x + 2)^2 -1 = 8
>> (3x + 2)^2 = 9
>> (3x + 2) = +-3
If 3x+2 = 3
>> 3x=1
>> x = 1/3
OR if 3x+2 = -3
>> 3x = -5
>> x = -5/3