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7nadin3 [17]
1 year ago
13

Solve √x-1= x-3. Check for extraneous solutions.

Mathematics
1 answer:
Semenov [28]1 year ago
5 0

Step-by-step explanation:

so, "- 1" is also part of the square root ?

sqrt(x - 1) = x - 3

it is clear that for any value x < 1 we have no solution in R (as this makes the argument of the square root negative, and there is so real number solution for the square root of negative numbers).

now square the whole equation.

x - 1 = (x - 3)² = x² - 6x + 9

x² - 7x + 10 = 0

the general solution for quadratic equations is

x = (-b ± sqrt(b² - 4ac))/(2a)

in our case

a = 1

b = -7

c = 10

x = (7 ± sqrt(49 - 4×1×10))/(2×1) =

= (7 ± sqrt(49 - 40))/2 = (7 ± sqrt(9))/2

x1 = (7 + 3)/2 = 10/2 = 5

x2 = (7 - 3)/2 = 4/2 = 2

x2 is probably (given the answer options) not a valid solution for the original problem, as it represents the negative solution of sqrt(x - 1).

sqrt(2 - 1) = 2 - 3

± 1 = -1

remember, every square root has always 2 solutions : a positive and a negative one.

your teacher clearly only wanted the positive solution, which is x1 = 5.

so, yes,

C. x = 5

is the correct answer.

but please send your teacher my regards and comments. he/she has to state that only the positive solution to the square root is required/allowed.

because, formally, also x = 2 is a valid solution.

and therefore, C. AND D. are correct answers !!!!

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