Is solution ok?

• Feb 15th 2006, 04:07 AM
DenMac21
Is solution ok?
I have to solve this irrational inequality:
$\displaystyle \sqrt {25 - x^2 } + \sqrt {x^2 + 7x} > 3$

Is this solution?
$\displaystyle \sqrt {25 - x^2 } > 0 \wedge \sqrt {x^2 + 7x} > 3$
$\displaystyle \sqrt {25 - x^2 } > 1 \wedge \sqrt {x^2 + 7x} > 2$
$\displaystyle \sqrt {25 - x^2 } > 2 \wedge \sqrt {x^2 + 7x} > 1$
$\displaystyle \sqrt {25 - x^2 } > 3 \wedge \sqrt {x^2 + 7x} > 0$
• Feb 15th 2006, 11:50 AM
earboth
Quote:

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

Is this solution?

Hello,

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

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

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

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

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

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

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

Best wishes

Bye