1. ## Is solution ok?

I have to solve this irrational inequality:
$\sqrt {25 - x^2 } + \sqrt {x^2 + 7x} > 3$

Is this solution?
$\sqrt {25 - x^2 } > 0 \wedge \sqrt {x^2 + 7x} > 3$
$\sqrt {25 - x^2 } > 1 \wedge \sqrt {x^2 + 7x} > 2$
$\sqrt {25 - x^2 } > 2 \wedge \sqrt {x^2 + 7x} > 1$
$\sqrt {25 - x^2 } > 3 \wedge \sqrt {x^2 + 7x} > 0$

2. Originally Posted by DenMac21
I have to solve this irrational inequality:
$\sqrt {25 - x^2 } + \sqrt {x^2 + 7x} > 3$

Is this solution?
Hello,

the bad news first: The right solution is: $x \in \{0 \leq x \leq 5\}$

The first root is defined for $x \in \{-5 \leq x \leq 5\}$

the second root is defined for $x \in \{x \leq -7 \vee x \geq 0\}$

that means the LHS of your inequality is defined for $x \in \{0 \leq x \leq 5\}$

For all x from 0 to 4 the first root is greater than 3, that means the inequality is true for $x \in \{0 \leq x \leq 4\}$

For all x greater than 4 the second root alone has a value greater than $\sqrt{44} \approx 6.6$

Therefore the inequality is true for all $x \in \{0 \leq x \leq 5\}$

Best wishes

Bye