Solve sqrt(x - 6) - sqrt(10 - x) >= 1

Question

Solve the inequality

\frac{7 + 16}{2} \leq x \leq 10

Alternative Form

x \in [\frac{7 + 16}{2}, 10]

Evaluate

x - 6 - 10 - x \geq 1

Find the domain

More Steps

x - 6 - 10 - x \geq 1, 6 \leq x \leq 10

Move the expression to the left side

x - 6 - 10 - x - 1 \geq 0

Move the expression to the right side

x - 6 \geq 10 - x + 1

Raise both sides of the inequality to the power of 2

x - 6 \geq (10 - x + 1)^{2}

Move the expression to the left side

x - 6 - (10 - x + 1)^{2} \geq 0

Calculate

More Steps

2 x - 17 - 2 10 - x \geq 0

Move the expression to the right side

- 2 10 - x \geq - 2 x + 17

Change the signs on both sides of the inequality and flip the inequality sign

2 10 - x \leq 2 x - 17

Separate the inequality into 2 possible cases

2 10 - x \leq 2 x - 17, 2 x - 17 \geq 0 2 10 - x \leq 2 x - 17, 2 x - 17 < 0

Solve the inequality

More Steps

x \in (- \infty, \frac{- 7 + 16}{2}] \cup [\frac{7 + 16}{2}, + \infty), 2 x - 17 \geq 0 2 10 - x \leq 2 x - 17, 2 x - 17 < 0

Solve the inequality

More Steps

x \in (- \infty, \frac{- 7 + 16}{2}] \cup [\frac{7 + 16}{2}, + \infty), x \geq \frac{17}{2} 2 10 - x \leq 2 x - 17, 2 x - 17 < 0

Since the left-hand side is always positive or 0,and the right-hand side is always negative,the statement is false for any value of x

x \in (- \infty, \frac{- 7 + 16}{2}] \cup [\frac{7 + 16}{2}, + \infty), x \geq \frac{17}{2} x \in \emptyset, 2 x - 17 < 0

Solve the inequality

More Steps

x \in (- \infty, \frac{- 7 + 16}{2}] \cup [\frac{7 + 16}{2}, + \infty), x \geq \frac{17}{2} x \in \emptyset, x < \frac{17}{2}

Find the intersection

x \geq \frac{7 + 16}{2} x \in \emptyset, x < \frac{17}{2}

Find the intersection

x \geq \frac{7 + 16}{2} x \in \emptyset

Find the union

x \geq \frac{7 + 16}{2}

Check if the solution is in the defined range

x \geq \frac{7 + 16}{2}, 6 \leq x \leq 10

Solution

\frac{7 + 16}{2} \leq x \leq 10

Alternative Form

x \in [\frac{7 + 16}{2}, 10]

Show Solution