√(5-x) = x+1
To be defined, 5-x ≥ 0
-x ≥ -5
x ≤ 5
also since √anything yields a positive result, x+1≥0
x≥-1
so -1 ≤ x ≤ 5
To solve, square both sides
5-x = x^2 + 2x + 1
x^2 + 3x - 4 = 0
(x+4)(x-1) = 0
x = -4 or x = 1
because of our domain, x = 1
Given: square root of 5 minus x is equals to x plus 1. Show that the solution to the above equation lies in the interval of negative 1less than or equals to x less than or equals to 5. Solve the equation. Solve the equation negative square root of 5 minus x is equals to x plus one
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