(5x-1)^2=4/25
5x-1=2/5 5x=7/5 x=7/25
right?
shouldn't it be: \(\large 5x-1=\pm \frac{2}{5} \) ???
Yes. we should include pm... ie plus minus..
ok so can you explain the step by step after sq rooting that you get 3/25 and 7/25 as the answer that is where i get lost
Take the square root both the sides and tell me what are you getting?
i'm getting 5x-1= 2/5, but what do i do after that?
b/c there are 2 answers 3/25 and 7/25
Yes, there will be two answers for this...
Actually, when we take square root of any number we consider plus minus sign with it : Like : \[(5x-1)^2 = \frac{4}{25}\] \[\implies 5x - 1 = \pm \frac{2}{5}\]
Now, firstly you will consider + sign with 2/5 and then in second case you will consider - sign with 2/5 Solve by using + 2/5 first..
For Plus Case : \[5x - 1 = \frac{2}{5}\] Solve this for x..
Now Minus Case : \[5x - 1 = -\frac{2}{5}\] And now find x in this case..
ok so then do i divided each side by 5?
Firstly add 1 both the sides...
\[5x - 1 = \frac{2}{5}\] Add 1 to both the sides and tell me what you got?
ok so then i have 5x= 1+2/5 = 5x=3/5
How you got 3/5 ??
\[1 + \frac{2}{5} = ?\]
i just added the 1 to the 2
No, you cannot add 1 to the fraction directly.. Do you know about LCM or LCD ??
oh yes
Then use that...
ok so the lcd would be 5 making the fraction 5/5 correct?
Correct..
ok so adding 5/5 to 2/5 gives you 7/10 am i wrong there
See, you, in LCD, never add denominator, the denominator remains the same.. Just add or subtract the numerator...
ok
\[\frac{5}{5} + \frac{2}{5} = \frac{5 + 2}{5} = \frac{7}{5}\]
thank you now i see where my issues were :)
So, now you got : \[5x = \frac{7}{5}\] Now divide by 5 here to get x..
you getr 7/25
Yes...
Now : \[5x - 1 = -\frac{2}{5}\] Add 1 to both the sides...
0.28,0.12
Join our real-time social learning platform and learn together with your friends!