which of the following do you mean
3n = √(10n) - 1
3n = √(10n-1)
3n = √10 n - 1
The way you typed it , it would be the last one, but somehow I doubt that is what you meant.
Please clarify the above, as well as your second question in this post.
3n=squareroot of 10n-1
And
Y=squareroot of 7y-10
Help would be greatly appreciated!
4 answers
Sorry yes it is the first one
The second would be that same way
The second would be that same way
then
3n = √(10n) - 1
3n + 1 = √(10n)
square both sides
9n^2 + 6n + 1 = 10n
9n^2 -4n + 1 = 0
This quadratic has no real roots.
Are you sure you did not mean
3n = √(10n-1) ????, because then after squaring
9n^2 = 10n - 1
9n^2 - 10n + 1 = 0
(n-1)(9n-1) = 0
n = 1 or n = 1/9
if n=1,
LS = 3
RS = √(10-1) = √9 = 3 = LS
if n=1/9
LS = 3(1/9) = 1/3
RS = √(10/9 - 1)
= √(1/9) = √1/√9 = 1/3 =
so n = 1 or n = 1/9
Do the second the same way after seeing it as
y = √(7y-10)
( I can see two "nice" answers)
3n = √(10n) - 1
3n + 1 = √(10n)
square both sides
9n^2 + 6n + 1 = 10n
9n^2 -4n + 1 = 0
This quadratic has no real roots.
Are you sure you did not mean
3n = √(10n-1) ????, because then after squaring
9n^2 = 10n - 1
9n^2 - 10n + 1 = 0
(n-1)(9n-1) = 0
n = 1 or n = 1/9
if n=1,
LS = 3
RS = √(10-1) = √9 = 3 = LS
if n=1/9
LS = 3(1/9) = 1/3
RS = √(10/9 - 1)
= √(1/9) = √1/√9 = 1/3 =
so n = 1 or n = 1/9
Do the second the same way after seeing it as
y = √(7y-10)
( I can see two "nice" answers)
Yep you were right. Thank you so much. I really appreciate it.