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alexandr402 [8]
3 years ago
7

John and Tim are looking at the equation the square root of the quantity of 3 times x minus 4 equals square root of x . John say

s that the solution is extraneous. Tim says the solution is non-extraneous. Is John correct? Is Tim correct? Are they both correct? Justify your response by solving this equation, explaining each step with complete sentences. (10 points)

Mathematics
1 answer:
IgorC [24]3 years ago
3 0
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)

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Suppose that we don't have a formula for g(x) but we know that g(3) = −5 and g'(x) = x2 + 7 for all x. (
True [87]
So it tells us that g(3) = -5 and g'(x) = x^2 + 7.

So g(3) = -5 is the point (3, -5)
Using linear approximation
g(2.99) is the point (2.99, g(3) + g'(3)*(2.99-3))

now we just need to simplify that
(2.99, -5 + (16)*(-.01)) which is (2.99, -5 + -.16) which is (2.99, -5.16)
So g(2.99) = -5.16 

Doing the same thing for the other g(3.01)
(3.01, g(3) + g'(3)*(3.01-3))
(3.01, -5 + 16*.01) which is (3.01, -4.84)
So g(3.01) = -4.84

So we have our linear approximation for the two. 

If you wanted to, you could check your answer by finding g(x).  Since you know g'(x), take the antiderivative and we will get 
g(x) = 1/3x^3 + 7x + C
Since we know g(3) = -5, we can use that to solve for C
1/3(3)^3 + 7(3) + C = -5 and we find that C = -35
so that means g(x) = (x^3)/3 + 7x - 35

So just to check our linear approximations use that to find g(2.99) and g(3.01)

g(2.99) = -5.1597
g(3.01) = -4.8397

So as you can see, using the linear approximation we got our answers as
g(2.99) = -5.16
g(3.01) = -4.84
which are both really close to the actual answer.  Not a bad method if you ever need to use it. 
5 0
3 years ago
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