There's actually a secret trick to doing it without using the complete the square method. There's a formula and it always works. My teacher taught me it and I used it a lot in the ALG1 EOC and was VERY helpful. I guess ur prepping for the eoc ri xD
So the "hack" is the first 2 attachments.
it's important to know the typical method, so the typical one is the last attachment
Answer:
it 1
Step-by-step explanation:
replace the log with 10 then log it
![{6x}^{2} - 24 = 0 \\ = > 6( {x}^{2} - 4) = 0 \\ = > 6 {((x)}^{2} - {(2)}^{2} ) = 0 \\ = > 6(x - 2)(x + 2) = 0 \\ = > 6(x - 2) = 0 \: \: \: or \: \: 6(x + 2) = 0 \\ = > 6x - 12 = 0 \: \: \: or \: \: \: 6x + 12 = 0 \\ = > x = 2 \: \: \: or \: \: \: x = - 2](https://tex.z-dn.net/?f=%20%7B6x%7D%5E%7B2%7D%20%20-%2024%20%3D%200%20%5C%5C%20%20%3D%20%20%3E%206%28%20%7Bx%7D%5E%7B2%7D%20%20-%204%29%20%3D%200%20%5C%5C%20%20%3D%20%20%3E%206%20%7B%28%28x%29%7D%5E%7B2%7D%20%20-%20%20%7B%282%29%7D%5E%7B2%7D%20%29%20%3D%200%20%5C%5C%20%20%3D%20%20%3E%206%28x%20-%202%29%28x%20%2B%202%29%20%3D%200%20%5C%5C%20%20%3D%20%20%3E%206%28x%20-%202%29%20%3D%200%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20or%20%5C%3A%20%20%5C%3A%206%28x%20%2B%202%29%20%3D%200%20%5C%5C%20%20%3D%20%20%3E%206x%20-%2012%20%3D%200%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20or%20%5C%3A%20%20%5C%3A%20%20%5C%3A%206x%20%2B%2012%20%3D%200%20%5C%5C%20%20%3D%20%20%3E%20x%20%3D%202%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20or%20%5C%3A%20%20%5C%3A%20%20%5C%3A%20x%20%3D%20%20-%202)
Hope you could get an idea from here.
Doubt clarification - use comment section.
The two numbers are 6 and 3
6+3=9
6(2)+3=15
Step-by-step explanation:
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