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 !!!!
