#Can someone give me the steps to solve these?
24 messages · Page 1 of 1 (latest)
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For the first one, all ya gotta do is solve for the values of x.
Subtract 4, square both sides, set equation equal to 0, then factor, then solve for x, and test for extraneous solutions by plugging in your solutions x in the original equation
For the second one, set it equal to 0, factor out GCF, set each factor equal to 0, solve for possible x-values
And boom ya get your answers
You cant just remove the 4
You need to write x-4 on the other side instead of x
And then square both sides
And when they said equate to zero, basically put all the terms to one side
Would it help if I send a pic of the steps?
Yes, sorry
Just remember to check for extraneous solutions
It is the case here that no solution is extraneous, but not always
You mean GCD?
The method I was taught was prime factorization method. I will illustrate this method by an example
12 and 18
We first prime factorize
12 = 2^2 * 3^1
18 = 2^1 * 3^2
To find GCD, pick the least amount of primes in the two numbers
we have 2^1 and 2^2. So least is 2^1
we have 3^1 and 3^2. least is 3^1
Now multiply them: 2 * 3 = 6
so 6 is the GCD of the numbers
There si another method called Euclidean Algorithm, but I don't have enough knowledge of it. You can search that
I meant Gcf, I'm going after the 2nd problem now and I'm this step
The gcf here 3
3x
@finite umbra
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