Inequalities and equations with roots
Task number: 2760
Solve the folowing inequalities and equations over the domain of real numbers:
Variant 1
\( \sqrt{x-2}+x>4 \)
Resolution
The expression is defined for \(x-2\ge 0 \longrightarrow x\ge 2\). We rewrite the inequality as \(\sqrt{x-2}> 4-x\). The left side of this inequality is always non-negative (as long as it is defined), so the inequality cannot hold whenever the right side is negative. That will happen for \(x\in (4,\infty)\). For \(2\leq x\leq 4\) both sides of the inequality are positive and so we can square them.
\(\sqrt{x-2}> 4-x \longrightarrow x-2> x^2-8x+16 \longrightarrow x^2-9x+18<0 \longrightarrow x_{1{,}2}=\frac{9\pm\sqrt{81-4{\cdot}18}}{2}=\frac{9\pm3}{2}\).
Result
The solution is \(x\in (3,\infty)\).
Variant 2
\( \sqrt{x^2+2x-3} \ge \sqrt{x^2+3x-4} \)
Resolution
The left side is defined for \(x^2+2x-3 \ge 0 \longrightarrow x_{1{,}2}=\frac{-2\pm\sqrt{4+12}}{2}=-1\pm 2 \longrightarrow \\ x\in (-\infty,-3\rangle \cup \langle 1,\infty)\).
The right side is defined for \(x^2+3x-4 \ge 0 \longrightarrow x_{1{,}2}=\frac{-3\pm\sqrt{9+16}}{2} \longrightarrow x\in (-\infty,-4\rangle \cup \langle 1,\infty)\).
Squaring both sides, we get \(x^2+2x-3 \ge x^2+3x-4 \longrightarrow x\le 1\).
The inequality holds when all three of these conditions are fulfilled simultaneously.
Result
The solution is \(x\in (-\infty,-4\rangle \cup \{1\}\).
Variant 3
\( \sqrt{x^2-1}\ge \sqrt{x^2+x-6} \)
Resolution
The left side is defined for \(x^2-1\ge 0 \longrightarrow x\in(-\infty,-1\rangle\cup\langle 1,\infty)\).
The right side is defined for \(x^2+x-6\ge 0 \longrightarrow x_{1{,}2}=\frac{-1\pm\sqrt{1+24}}{2}\longrightarrow x\in(-\infty,-3\rangle\cup\langle 2,\infty)\).
Squaring, we have \(x^2-1 \ge x^2+x-6 \longrightarrow x\le 5\).
Result
The solution is \(x\in(-\infty,-3\rangle\cup\langle 2{,}5\rangle\).
Variant 4
\( \sqrt{x+1}-\sqrt{x-4}=1 \)
Resolution
The first square root is defined for \(x+1\ge 0 \longrightarrow x\ge -1\).
The second square root is defined for \(x-4\ge 0 \longrightarrow x\ge 4\).
Squaring, we have \(x+1 +2\sqrt{(x+1)(x-4)}+x-4 =1 \longrightarrow \sqrt{x^2-3x-4}=x-2 \longrightarrow x^2-3x-4=x^2-4x+4 \longrightarrow x=8\).
Checking this solution, \(\sqrt{8+1}-\sqrt{8-4}=3-2=1\).
Result
The solution is \(x=8\).
