Disequation Probem

[petitpauline]

Hi i need some help or guide to solve this disequation :

Classify the disequation and then solve

Thank you

 

[soroban]

Hello, petitpauline!

\(\displaystyle \text{Solve the inequality: }\;\frac{2\sqrt{x+5}\,-\sqrt{15-x}}{\sqrt{x}-2} \;>\;0\)

\(\displaystyle \text{We see immediately that: }\;0 \,\le\, x \,\le\,15 \:\text{ and }\:x \ne 4\)


\(\displaystyle \text{A fraction is positive if :}\)

. . \(\displaystyle \text{(1) both the numerator and denominator are positive.}\)

. . \(\displaystyle \text{(2) both the numerator and denominator are negative.}\)


\(\displaystyle \text{(1) Both positive.}\)

\(\displaystyle \begin{array}{cccccccccc}2\sqrt{x +5} - \sqrt{15-x} \;>\;0 & \Rightarrow & 2\sqrt{x+5} \;>\;\sqrt{15-x} & \Rightarrow & 4(x+5)\:>\:15-x & \Rightarrow& x \:>\:-1 \\ \\ \sqrt{x} - 2 \;>\;0 & \Rightarrow & \sqrt{x} \:>\:2 & \Rightarrow & x \;>\;4 \end{array}\)

. . \(\displaystyle \text{Hence: }\;4\;<\;x\;\le\;15\)


\(\displaystyle \text{(2) Both negativee.}\)

\(\displaystyle \begin{array}{cccccccccc}2\sqrt{x +5} - \sqrt{15-x} \;<\;0 & \Rightarrow & 2\sqrt{x+5} \;<\;\sqrt{15-x} & \Rightarrow & 4(x+5)\:<\:15-x & \Rightarrow& x \:<\:-1 \\ \\ \sqrt{x} - 2 \;<\;0 & \Rightarrow & \sqrt{x} \:<\:2 & \Rightarrow & x \;<\;4 \end{array}\)

. . \(\displaystyle \text{Hence: }\;0\le \;x\;<\;4\)


\(\displaystyle \text{Therefore: }\;x \:\in\:\left[\;\! 0,4)\cup(4,15 \;\!\right]\)

 

[petitpauline]

Thank you
i' ve finally understood

 

[mmm4444bot]

i' ve finally understood

I don't think so.

\(\displaystyle \text{We see immediately that: }\;0 \,\le\, x \,\le\,15 \:\text{ and }\:x \ne 4\)

I'm thinking that Soroban intends this beginning information to be a domain statement for the rational expresson on the lefthand side, rather than a solution statement for the inequality itself.

Soroban's domain statement is correct, but he suffered a minor lapse in logic (as we ALL do, from time to time) at the end.

It seems that you (petitpauline) missed the error, so your understanding is not yet complete, for this exercise.

In Case II (where the sign of the numerator and denominator are both negative), Soroban correctly shows the constraints on x:

x < -1 AND x < 4

This conclusion tells us that x must be smaller than -1.

But, we've already seen that the rational expression does not exist for negative values of x.

Therefore, no solution to the inequality comes from Case II.

The solution to the posted exercise is 4 < x ? 15.

Cheers