You can only use the symbols and numbers once and you have to use every number but not all symbols... please answer this first o
ne gets a 100 points
2 answers:
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The equation is
3x - 4 = sqrt(x)
where 'sqrt' is shorthand for 'square root'
Let's solve the equation. To do so, we square both sides. Then we get everything to one side
3x - 4 = sqrt(x)
(3x - 4)^2 = (sqrt(x))^2
9x^2 - 24x + 16 = x
9x^2 - 24x + 16-x = x-x
9x^2 - 25x + 16 = 0
Now use the quadratic formula. See the attached image for those steps.
After using the quadratic formula the two possible solutions are x = 1 or x = 16/9
We need to check each possible solution to see if it is extraneous or not.
----------------------------
Checking x = 1
3x - 4 = sqrt(x)
3*1 - 4 = sqrt(1)
3 - 4 = 1
-1 = 1
The final equation is false, so x = 1 is not a true solution
x = 1 is extraneous.
So far, John is correct; however, we need to see the nature of the possible solution x = 16/9
So let's check it
----------------------------
Checking x = 16/9
3x - 4 = sqrt(x)
3*(16/9) - 4 = sqrt(16/9)
48/9 - 4 = 4/3
16/3 - 4 = 4/3
16/3 - 12/3 = 4/3
(16 - 12)/3 = 4/3
4/3 = 4/3
The last equation is true, so x = 16/9 is a proper solution.
This solution is considered non-extraneous. So Tim is also correct
----------------------------
They are both correct because there are two possible solutions. One of which is extraneous (x = 1) and the other is non-extraneous (the fraction x = 16/9)
3x - 4 = sqrt(x)
where 'sqrt' is shorthand for 'square root'
Let's solve the equation. To do so, we square both sides. Then we get everything to one side
3x - 4 = sqrt(x)
(3x - 4)^2 = (sqrt(x))^2
9x^2 - 24x + 16 = x
9x^2 - 24x + 16-x = x-x
9x^2 - 25x + 16 = 0
Now use the quadratic formula. See the attached image for those steps.
After using the quadratic formula the two possible solutions are x = 1 or x = 16/9
We need to check each possible solution to see if it is extraneous or not.
----------------------------
Checking x = 1
3x - 4 = sqrt(x)
3*1 - 4 = sqrt(1)
3 - 4 = 1
-1 = 1
The final equation is false, so x = 1 is not a true solution
x = 1 is extraneous.
So far, John is correct; however, we need to see the nature of the possible solution x = 16/9
So let's check it
----------------------------
Checking x = 16/9
3x - 4 = sqrt(x)
3*(16/9) - 4 = sqrt(16/9)
48/9 - 4 = 4/3
16/3 - 4 = 4/3
16/3 - 12/3 = 4/3
(16 - 12)/3 = 4/3
4/3 = 4/3
The last equation is true, so x = 16/9 is a proper solution.
This solution is considered non-extraneous. So Tim is also correct
----------------------------
They are both correct because there are two possible solutions. One of which is extraneous (x = 1) and the other is non-extraneous (the fraction x = 16/9)
Step-by-step explanation:
area of square = 1
area of a circle in square =
bu estimate
b) if n tend to infinity