Find the Domain x+5 square root of x^2+11x+3

Problem

x+5√(,x2+11*x+3)

Solution

1. Identify the restriction for the square root function. For the expression to be defined in the set of real numbers, the radicand (the expression inside the square root) must be greater than or equal to zero.

x2+11*x+3≥0

2. Find the roots of the quadratic equation x2+11*x+3=0 using the quadratic formula x=(−b±√(,b2−4*a*c))/(2*a)

x=(−11±√(,11−4*(1)*(3)))/(2*(1))

3. Simplify the expression under the radical to determine the critical points.

x=(−11±√(,121−12))/2

x=(−11±√(,109))/2

4. Determine the intervals where the quadratic x2+11*x+3 is non-negative. Since the parabola opens upward (the coefficient of x2 is positive), the expression is greater than or equal to zero outside the roots.

x≤(−11−√(,109))/2* or *x≥(−11+√(,109))/2

5. Express the domain in interval notation.

(−∞,(−11−√(,109))/2]∪[(−11+√(,109))/2,∞)

Final Answer

Domain=(−∞,(−11−√(,109))/2]∪[(−11+√(,109))/2,∞)

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