How to solve irrational equations and inequalities?

• Example 1 - Solve the equation \(\sqrt{2x+3} = 2.\)
  Solution - The condition is that everything underneath the root must be non negative value. $$\begin{eqnarray}2x+3 &\geq& 0\\ \nonumber 2x &\geq& -3/:2 \\ \nonumber x &\geq& -\frac{3}{2}\end{eqnarray}$$ So the condition is that x must be equal or greater than -3/2 (-1.5) so that the term under the root would be non negative. $$ \begin{eqnarray}\sqrt{2x+3} &=& 2/^2\\ \nonumber 2x +3 &=& 4 \\ \nonumber 2x &=& 1 /:2 \\\nonumber x &=& \frac{1}{2}\end{eqnarray}$$ The obtained solution is in accordnace with the condition. Proof that solution is correct. $$\begin{eqnarray}\sqrt{2\cdot \frac{1}{2} + 3} &=& 2\\ \nonumber \sqrt{4} &=& 2 \\\nonumber 2 &=& 2\end{eqnarray}$$
• Example 2 - Solve the equation \(\sqrt{2x+3} = -4.\)
  Solution - This equation do not have solution since the left size of the equation is greather or equal to zero. Proof: $$ \begin{eqnarray} 2x+3 &\geq& 0 \\\nonumber 2x&\geq& -3 \\\nonumber x&\geq& -\frac{3}{2}\end{eqnarray}$$ As seen from the previously defined condition the x must be greather or equal to \(\frac{3}{2}.\) However, in the original inequality the term on the left side is equal to -2 which is lower value than in obtained condition.
• Example 3 - Solve the equation \(\sqrt{x^2 + x + 1} = x-1\)
  Solution - In this equation the left and the right hand side of the equation must be greather or equal to 0. $$\begin{eqnarray}\sqrt{x^2+x+1}&=& x-1/^2 \\ \nonumber x^2 + x + 1 &=& (x-1)^2 \\\nonumber x^2 + x + 1 &=& x^2 - 2x + 1 \\ \nonumber 3x &=& 0 /:3 \\ \nonumber x &=& 0\end{eqnarray}$$
• Example 4 - Solve the equation \(\sqrt{x+1} + \sqrt{x+2} = 1\)
  Solution - Before solving the equation the terms under square roots must be greater or equal to zero. $$ x + 1 \geq 0 \Rightarrow x \geq -1, $$ $$ x+2 \geq 0 \Rightarrow x \geq -2$$ Now the equation can be solved. $$\begin{eqnarray}\sqrt{x+1} &=& 1- \sqrt{x+2}/^2\\\nonumber x+1 &=& 1 - 2\sqrt{x+2} + x+2 \\\nonumber 2\sqrt{x+2} &=& 2/:2 \\\nonumber \sqrt{x+2} &=& 1/^2\\\nonumber x+2 &=& 1 \\\nonumber x &=& -1\end{eqnarray}
• Example 5 - Solve the equation \(\sqrt{3-\sqrt{x-3}}=1.\)
  Solution$$ \begin{eqnarray}\sqrt{3-\sqrt{x-3}} &=& 1/^2 \\\nonumber 3 - \sqrt{x-3}&=&1/\cdot(-1)\\\nonumber \sqrt{x-3} &=& 2\end{eqnarray}$$ Before solving the equation the expression under square root must be greater or equal to 0. $$x-3 \geq 0 \Rightarrow x \geq 3$$ $$\begin{eqnarray}\sqrt{x-3} &=& 2/^2 \\\nonumber x-3 &=& 4 \\\nonumber x &=& 7\end{eqnarray}$$ The condition is satisfied due to the fact that \(x\) is greater than 3.
• Example 6 - Solve the inequality \(\sqrt{x-3} > 9\)
  Solution - Before solving the equation the expression under square root must be greater or equal to 0. \(x-3\geq 0 \Rightarrow x \geq 3.\) When the inequality is squared the following solution is obtained. $$\begin{eqnarray}\sqrt{x-3}&\geq& 9 /^2 \\ \nonumber |x-3|&\geq& 81\end{eqnarray}$$ From the obtained expression two inequalities emerge. \( x-3 > 81,\) and \(-x-3 > 81 \). The solution of the first inequality. $$\begin{eqnarray}x-3 &>& 81 \\\nonumber x&>& 84\end{eqnarray}$$ The solution of the second inequality. $$\begin{eqnarray}-(x-3)&>&81/\cdot(-1)\\\nonumber x-3 & <& -81 \\\nonumber x &<&-78\end{eqnarray}$$ The second condition is omitted due to the fact that variable \(x\) must be greater than 3. So the solution can be written as \(x \in \langle 84, +\infty\rangle\)
• Example 7 - Solve the inequality \(\sqrt{x-3} > 3\)
  Solution \begin{eqnarray}\sqrt{x-3} &>& 3/^2\Rightarrow x-3 \geq 0 \Rightarrow x \geq 3\end{eqnarray}
• Example 8 - Solve the equation \(\sqrt{x^2-4} = 1\)
  Solution - The condition is that expression under the square root must be equal or greater than 0. $$\begin{eqnarray} x^2 - 4 \geq 0 \Rightarrow x^2 \geq 4 /^\sqrt{} \Rightarrow x_{1,2} \geq \pm 2 \\\nonumber x \in \langle -\infty, -2] \cup [2, +\infty\rangle\end{eqnarray} $$ The solution of the equation can be written as $$\begin{eqnarray}\sqrt{x^2-4} &=& 1/^2 \\\nonumber x^2-4 &=& 1\\ \nonumber x^2 &=& 5 /^\sqrt{} \\\nonumber x_{1,2} &=& \pm \sqrt{5}\end{eqnarray}$$
• Example 9 - Solve the equation \(\sqrt{x+10} - \sqrt{x+7} = 1\)
  Solution - Before the equation is solved the conditions must be defined i.e. the expressions under square roots must be greater or equal to 0. $$ x + 10 \geq 0 \Rightarrow x \geq -10,$$ $$ x + 7 \geq 0 \Rightarrow x \geq -7$$ The solution of the equation can be written as $$\begin{eqnarray}\sqrt{x+10} - \sqrt{x+7} &=& 1\\\nonumber \sqrt{x+10} &=& 1 + \sqrt{x+7}/^2 \\\nonumber x+10 &=& 1 +2 \sqrt{x+7} + x+7 \\\nonumber 2 \sqrt{x+7} &=& 2 /:2 \\ \nonumber \sqrt{x+7} &=& 1 /^2 \\ \nonumber x +7 &=& 1 \\\nonumber x &=& - 6\end{eqnarray}$$
• Example 10 - Calculate the sum of the squares of the solution of the equation \(\sqrt{2x+5} - \sqrt{x+1} = 2\)
  Solution - The expressions under square roots must be greater or equal to 0. $$ 2x +5 \geq 0 \Rightarrow x \geq -\frac{5}{2}$$ $$ x+1 \geq 0 \Rightarrow x \geq -1 $$ The solution of the equation can be written as $$\begin{eqnarray}\sqrt{2x+5} - \sqrt{x+1} &=& 2 \\ \nonumber \sqrt{2x+5} &=& 2 + \sqrt{x+1} /^2 \\\nonumber 2x+5 &=& 4 + 4 \sqrt{x+1} + x + 1\\\nonumber 2x +5 -4-x-1 &=& 4\sqrt{x+1}\\ \nonumber x &=& 4\sqrt{x+1}/^2 \\\nonumber x^2 &=& 16(x+1) \\\nonumber x^2 -16x -16 &=& 0 \\\nonumber\end{eqnarray}$$ The next step is to solve the quadratic equation. $$\begin{eqnarray}x_{1,2} &=& \frac{16 \pm \sqrt{16^2 + 4\cdot 16}}{2}\end{eqnarray}$$ $$ \begin{eqnarray}x_{1,2} &=& \frac{16 \pm \sqrt{256+64}}{2}\\\nonumber x_{1,2} &=& \frac{16 \pm \sqrt{320}}{2} \\\nonumber x_{1,2} &=& \frac{16 \pm 8\sqrt{5}}{2}\\\nonumber x_{1,2} &=& 8 \pm 4 \sqrt{5} \\\nonumber x_1 &=& 8 + 4\sqrt{5} \\\nonumber x_2 &=& 8 - 4 \sqrt{5}\end{eqnarray}$$ The sum of the squares of the equation solution can be written as $$ \begin{eqnarray}x_1^2 + x_2^2 &=& (8+4\sqrt{5})^2 + (8 - 4\sqrt{5})^2\\\nonumber x_1^2 + x_2^2 &=& 64 + 2\cdot 16 \cdot 4\sqrt{5} + 80 + 64 - 2\cdot 16 \cdot 4\sqrt{5} + 80 \\\nonumber x_1^2 + x_2^2 &=& 128 + 160\\\nonumber x_1^2+x_2^2 &=& 288\end{eqnarray}$$
• Example 11 - Solve the equation \(\sqrt{x+1} - \sqrt{x-4} = 1\).
  Solution - The expressions under the square roots have to be greater than or equal to 0. $$\begin{eqnarray} x - 1 &\geq& 0 \Rightarrow x \geq 1\\ \nonumber x - 4 &\geq& 0 \Rightarrow x \geq 4\end{eqnarray}$$ The solution of the equation can be written as $$\begin{eqnarray} \sqrt{x+1} &=& 1 + \sqrt{x-4}/^2\\\nonumber x+1 &=& 1 + 2\sqrt{x-4} + x-4 \\\nonumber 2\sqrt{x-4} &=& 4/:2 \\\nonumber \sqrt{x-4} &=& 2 /^2 \\\nonumber x-4 &=& 4\\\nonumber x &=& 8\end{eqnarray}$$
• Example 12 - Solve the equation \(\sqrt{3x+7} - \sqrt{x+1} = 2\), and calculate the difference between solutions.
  Solution - The expressions under square roots must be greater than or equal to 0. $$\begin{eqnarray}3x+7&\geq& 0 \Rightarrow 3x&\geq& -7/:3 \Rightarrow x &=& -\frac{7}{3}\\\nonumber x+1 &\geq& 0 \Rightarrow x \geq -1\end{eqnarray}$$ The solution of the equation can be written as: $$\begin{eqnarray}\sqrt{3x+7} &=& 2 + \sqrt{x+1}/^2\\\nonumber 3x+7 &=& 4 + 4\sqrt{x+1} + x + 1\\\nonumber 2x +2 &=& 4 \sqrt{x+1}/:2 \\\nonumber x + 1&=& 2 \sqrt{x+1}/^2 \\\nonumber x^2 + 2x + 1 &=& 4x+4\\\nonumber x^2 -2x -3 &=& 0 \\ \nonumber\end{eqnarray} $$ The solution of quadratic equation can be written as $$\begin{eqnarray}x_{1,2} &=& \frac{2\pm \sqrt{4 + 4\cdot 3}}{2}\\\nonumber x_{1,2} &=& \frac{2 \pm 4}{2}\\\nonumber x_{1,2} &=& 1\pm 2\\\nonumber x_1 &=& 1 + 2 = 3 \\\nonumber x_2 &=& 1 -2 = -1 \end{eqnarray}$$ The sum of obtained solutions \(x_1 + x_2 \) is equal to 2.
• Example 13 - Solve the equation \(x^2 + 2x = 3x\sqrt{x}\) and calculate the multiplication of equation solutions.
  Solution - $$\begin{eqnarray}x(x+2) &=& 3x\sqrt{x}/:x \Rightarrow x_1 = 0\\\nonumber x+2 &=& 3\sqrt{x}/^2 \\ \nonumber x^2 + 4x + 4 &=& 9x \\\nonumber x^2 - 5x + 4 &=& 0\\\nonumber\end{eqnarray}$$ The solution of quadratic equation: \begin{eqnarray}x_{2,3}&=& \frac{5 \pm \sqrt{25 - 4\cdot 4}}{2}\\\nonumber x_{2,3} &=& \frac{5 \pm 3}{2}\\ \nonumber x_2 &=& \frac{5+3}{2} = \frac{8}{2} = 4 \\\nonumber x_3 &=& \frac{5-3}{2} =\frac{2}{2} = 1\end{eqnarray} Multiplication of equation solutions \(x_1\cdot x_2 \cdot x_3 = 0 \cdot 2 \cdot 1 = 0\)
• Example 14 - Solve the equation \(\sqrt{12 + 3\sqrt{x+1}} = 4\)
  Solution - \begin{eqnarray}\sqrt{12 + 3\sqrt{x+1}} &=& 4/^2\\\nonumber 12 + 3 \sqrt{x+1} &=& 16 \\ \nonumber 3\sqrt{x+1} &=& 4/^2 \Rightarrow x \geq -1 \\\nonumber 9x + 9 &=& 16\\\nonumber 9x &=& 7 \\\nonumber x&=& \frac{7}{9}\end{eqnarray} The solution of the equation\(\left(\frac{7}{9}\right)\) is much greater than the minimum value defined in the condtion for variable x.
• Example 15 - Solve the inequality \(\sqrt{3-x} > 3\)
  Solution - Before solving the inequality the expression under the square root must be greater or equal to zero. $$3-x \geq 0 \Rightarrow -x \geq -3 /(-1) \Rightarrow x \leq 3$$ \begin{eqnarray}\sqrt{3-x} &>& 3 /^2 \\\nonumber|3-x| &>& 9 \end{eqnarray} Solution of the first inequality \begin{eqnarray}3-x &>& 9 \\\nonumber x &<& 6\end{eqnarray} Solution of thesecond inequality \begin{eqnarray}-(3-x)&>& 9/\cdot(-1)\\\nonumber 3-x &<& -9 \\\nonumber -x &<& -12 /\cdot (-1)\\\nonumber x &>& 12\end{eqnarray}
• Example 16 - Solve the inequality \(\sqrt{4x-5} \leq 5\)
  Solution - The expression under the square root must be greater than or equal to 0. $$4x-5 \geq 0 \Rightarrow 4x \geq 5 /:4 x \geq \frac{5}{4}$$ \begin{eqnarray}\sqrt{4x-5} &\leq& 5/^2 \\\nonumber |4x-5| &\leq& 25 \end{eqnarray} There are two possible solutions to this inequality. The first solution \begin{eqnarray}4x-5 &\leq& 25\\\nonumber 4x &\leq& 30/:4 \\\nonumber x &\leq& 7.5 \end{eqnarray} The second solution of the inequlaity \begin{eqnarray}-(4x-5) &\leq& 25 /\cdot(-1) \\\nonumber 4x-5 &\geq& -25 \\\nonumber 4x &\geq& -20/:4 \\\nonumber x &\geq& -5\end{eqnarray}
• Example 17 - Solve the equation \(\sqrt{6+\sqrt{x-3}} = 3\)
  Solution - \begin{eqnarray}\sqrt{6+\sqrt{x-3}} &=& 3/^2 \\\nonumber 6 + \sqrt{x-3} &=& 9 \\\nonumber \sqrt{x-3} &=& 3/^2 \Rightarrow x-3 \geq 0 \Rightarrow x \geq 3\\\nonumber x-3 &=& 9 \\\nonumber x &=& 12\end{eqnarray}
• Example 18 - Solve the equation \(\sqrt{x-1} = 1\)
  Solution - \begin{eqnarray}\sqrt{x-1} &=& 1/^2 \Rightarrow x-1 \geq 0 \Rightarrow x \geq 1 \\\nonumber x-1 &=& 1\\\nonumber x &=& 2\end{eqnarray}
• Example 19 - Solve the equation \(\sqrt{x^2-36} = 8\)
  Solution - \begin{eqnarray}\sqrt{x^2-36} &=& 8 /^2 \Rightarrow x^2-36 \geq 0 \Rightarrow x^2 \geq 36 /^\sqrt{} \Rightarrow x_{1,2} \geq \pm 6 \\\nonumber x^2-36 &=& 64\\\nonumber x^2 &=& 100 /^\sqrt{} \\\nonumber x_{1,2} &=& \pm 10 \end{eqnarray}
• Example 20 - Solve the equation \(4 + \sqrt{x^2 +4} = x\)
  Solution - \begin{eqnarray}\sqrt{x^2 +4} &=& x-4/^2 \\\nonumber x^2 + 4 &=& x^2 - 8x + 16\\\nonumber 8x &=& 12 \\\nonumber x &=& \frac{12}{8} \\\nonumber x&=& \frac{3}{2}\end{eqnarray}
• Example 21 - Solve the eqution \(x+\sqrt{x^2+5} = 5.\)
  Solution - \begin{eqnarray}\sqrt{x^2+5} &=& 5- x/^2\\\nonumber x^2+5 &=& 25 - 10x + x^2 \\\nonumber 10x &=& 20 /:10 \\\nonumber x &=& 2\end{eqnarray}
• Example 22 - Solve the equation \(1+ \sqrt{5x+9} = x\)
  Solution - \begin{eqnarray}\sqrt{5x+9} &=& x-1/^2 \Rightarrow 5x + 9 \geq 0 \Rightarrow x \geq - \frac{9}{5}\\\nonumber 5x+9 &=& x^2 -2x +1 \\\nonumber -x^2 +7x +8 &=& 0/\cdot(-1)\\\nonumber x^2 -7x - 8 &=& 0 \\\nonumber x_{1,2} &=& \frac{7\pm\sqrt{49 + 32}}{2} \\\nonumber x_{1,2} &=& \frac{7\pm 9}{2} \\\nonumber x_{1} &=& \frac{7+9}{2} = \frac{16}{2} =8 \\\nonumber x_2 &=& \frac{7-9}{2} = -\frac{2}{2} = -1\end{eqnarray}
• Example 23 - Solve the equation \(\frac{\frac{1-2\sqrt{x}}{2-3\sqrt{x}}}{\frac{3-4\sqrt{x}}{4-5\sqrt{x}}} = \frac{5}{6}\)
  Solution - \begin{eqnarray}\frac{\frac{1-2\sqrt{x}}{2-3\sqrt{x}}}{\frac{3-4\sqrt{x}}{4-5\sqrt{x}}} &=& \frac{5}{6}\\\nonumber \frac{(1-2\sqrt{x})(4-5\sqrt{x})}{(2-3\sqrt{x})(3-4\sqrt{x})} &=& \frac{5}{6}\\\nonumber \frac{4 - 5\sqrt{x} - 8 \sqrt{x} + 10x}{6-8\sqrt{x}-9\sqrt{x}12x} &=& \frac{5}{6}\\\nonumber \frac{4-13\sqrt{x}+10x}{6-17\sqrt{x} + 12x} &=& \frac{5}{6} \\\nonumber 24-78\sqrt{x} + 60x &=& 30-85\sqrt{x}+60x\\\nonumber 7\sqrt{x} &=& 6/:7/^2\\\nonumber x &=& \frac{36}{49}\end{eqnarray}
• Example 24 - Solve the equation \(\sqrt{4x+13} + \sqrt{x+1} = \sqrt{3x+12}\)
  Solution - \begin{eqnarray}\sqrt{4x+13} + \sqrt{x+1} &=& \sqrt{3x+12}/^2\\\nonumber 4x + 13 + 2\sqrt{(4x+13)(x+1)} + x + 1 &=& 3x+12 \\\nonumber 2\sqrt{(4x+13)(x+1)} &=& -2x-2/:2\\\nonumber \sqrt{4x+13)(x+1)} &=& -(x+1)/^2\\\nonumber (4x+13)(x+1) &=& x^2 + 2x + 1\\\nonumber 4x^2 + 4x + 13x + 13 &=& x^2 + 2x + 1\\\nonumber 3x^2 +15x + 12 &=& 0/:3 \\\nonumber x^2 + 5x + 4 &=& 0 \\\nonumber x_{1,2} &=& \frac{-5 \pm \sqrt{25-4\cdot 4}}{2} = \frac{-5 \pm 3}{2}\\\nonumber x_1 &=& \frac{-5+3}{2} =-1 \\\nonumber x_2 &=& \frac{-5-3}{2} = -4\end{eqnarray}
• Example 25 - Solve the equation \(\sqrt{x-2} + \sqrt{4-x} = \sqrt{6-x}\)
  Solution - \begin{eqnarray}\sqrt{x-2} + \sqrt{4-x} &=& \sqrt{6-x}/^2 \\\nonumber x-2 +2 \sqrt{(x-2)(4-x)} + 4-x &=& 6-x \\\nonumber 2\sqrt{(x-2)(4-x)} &=& -x + 4/^2\\\nonumber 4(x-2)(4-x) &=& (4-x)^2\\\nonumber 4(4x-x^2-8+2x) &=& 16- 8x + x^2\\\nonumber 16x-4x^2-32+8x &=& 16-8x+x^2\\\nonumber -5x^2+32x-48 &=& 0/\cdot(-1)\\\nonumber 5x^2 -32x+48 &=& 0\\\nonumber x_{1,2} &=& \frac{32 \pm \sqrt{32^2 - 4\cdot 5 \cdot 48}}{10} \\\nonumber x_{1,2} &=& \frac{32 \pm \sqrt{64}}{10}\\\nonumber x_{1,2} &=& \frac{32 \pm 8}{10} \\\nonumber x_1 &=& \frac{40}{10} = 4 \\\nonumber x_2 &=& \frac{24}{10}=\frac{12}{5}\end{eqnarray}
• Example 26 - Solve the equation \(\sqrt{4x+8} = 2+\sqrt{3x-2}\)
  Solution - \begin{eqnarray}\sqrt{4x+8} &=& 2+\sqrt{3x-2}/^2 \\\nonumber 4x+8 &=& 4 + 4\sqrt{3x-2} + 3x-2 \\\nonumber x+6 &=& 4 \sqrt{3x-2}/^2\\\nonumber x^2 + 12x +36 &=& 16(3x-2)\\\nonumber x^2 +12x +36 &=& 48x - 32 \\\nonumber x^2 -36x + 68 &=& 0 \\\nonumber x_{1,2}&=& \frac{36\pm \sqrt{1024}}{2} \\\nonumber x_{1,2}&=& \frac{36\pm32}{2} \\\nonumber x_1 &=& \frac{36 + 32}{2} = \frac{68}{2} = 34\\\nonumber x_2 &=& \frac{36-32}{2} = \frac{4}{2} = 2\end{eqnarray}
• Example 27 - Solve the equation \(2(x-3) - \sqrt{x+2} = 0\)
  Solution - \begin{eqnarray}2(x-3) - \sqrt{x+2} &=& 0 \\\nonumber 2(x-3) &=& \sqrt{x+2}/^2 \\\nonumber 4(x^2 -6x + 9) &=& x+2 \\\nonumber 4x^2 -24 x + 36 &=& x+2 \\\nonumber 4x^2 -25x + 34 &=& 0 \\\nonumber x_{1,2} &=& \frac{25\pm \sqrt{625-4\cdot 4 \cdot 34}}{8} \\\nonumber x_{1,2} &=& \frac{25\pm \sqrt{81}}{8} \\\nonumber x_{1,2} &=& \frac{25 \pm 9}{8}\\\nonumber x_1 &=& \frac{34}{8} = \frac{17}{4} \\\nonumber x_2 &=& \frac{16}{8}=2\end{eqnarray}
• Example 28 - Solve the equation \(\sqrt{x+27} =2 +\sqrt{x-5}.\)
  Solution - \begin{eqnarray}\sqrt{x+27} &=& 2 +\sqrt{x-5}/^2\\\nonumber x+27 &=& 4 + 4\sqrt{x-5} + x-5\\\nonumber 4\sqrt{x-5} &=& 28 /:4\\\nonumber \sqrt{x-5} &=& 7/^2\\\nonumber x-5 &=& 49 \\\nonumber x &=& 54\end{eqnarray}
• Example 29 - Solve the equation \(\sqrt{x+11} - \sqrt{x+6} = \sqrt{4x+33}.\)
  Solution - \begin{eqnarray}\sqrt{x+11} - \sqrt{x+6} &=& \sqrt{4x+33}/^2\\\nonumber x+11 -2\sqrt{(x+11)(x+6)} + x+6 &=& 4x+11 \\\nonumber 2x +17 + 2 \sqrt{(x+11)(x+6)} &=& 4x+33 \\\nonumber 2\sqrt{(x+11)(x+6)} &=& 2x+ 16/:2\\\nonumber \sqrt{(x+11)(x+6)} &=& x+ 8/^2 \\\nonumber x^2 + 6x + 11x +66 &=& x^2 + 16x + 64 \\\nonumber 17x - 16x &=& 64-66 \\\nonumber x&=&-2 \end{eqnarray}
• Example 30 - Solve the equation \(\sqrt{9x+40} = 5\sqrt{x+5} - \sqrt{4x+5}\)
  Solution - \begin{eqnarray}\sqrt{9x+40} &=& 5\sqrt{x+5} - \sqrt{4x+5}/^2 \\\nonumber 9x+40 &=& 25(x+3) - 10\sqrt{(x+3)(4x+5)} + 4x+5\\\nonumber 9x+40 &=& 25x+75 -10\sqrt{(x+3)(4x+5)} + 4x+5\\\nonumber 20x + 40 &=& 10\sqrt{(x+3)(4x+5)}/^2 \\\nonumber 2x+4 &=& \sqrt{(x+3)(4x+5)}/^2 \\\nonumber 4x^2 + 16 x + 16 &=&(x+3)(4x+5)\\\nonumber 4x^2 +16x + 16 &=& 4x^2 +5x + 12x + 15 \\\nonumber -x&=&-1 \\\nonumber x &=& 1\end{eqnarray}
• Example 31 - Solve the equation \(\sqrt{3x+1} + \sqrt{x+4} = \sqrt{9-x}\)
  Solution - \begin{eqnarray}\sqrt{3x+1} + \sqrt{x+4} &=& \sqrt{9-x}/^2\\\nonumber 3x+1 + 2\sqrt{(3x+1)(x+4)} + x+4 &=& 9-x \\\nonumber 2\sqrt{(3x+1)(x+4)} &=& 4 - 5x /^2 \\\nonumber 4(3x+1)(x+4) &=& 16-40x+25x^2 \\\nonumber 4(3x^2 +12x+ x + 4) &=& 16 - 40x + 25x^2 \\\nonumber 12x^2 +48x +4x +16 &=& 16 - 40x + 25x^2 \\\nonumber -13x^2 + 92x &=& 0\\\nonumber x(92-13x)&=& 0\\\nonumber x_1 &=& 0 \\\nonumber -13x_2 &=& -92 \\\nonumber x_2&=& \frac{92}{13}\end{eqnarray}
• Example 32 - Solve the inequality \(\sqrt{2x+3} \leq 5\)
  Solution - The expression under root must be greater than or equal to 0. \(2x+3\geq 0 \Rightarrow 2x \geq -3/:2 \Rightarrow x\geq -\frac{3}{2}.\) \begin{eqnarray}\sqrt{2x+3} &\leq& 5/^2 \\\nonumber |2x+3| &\leq& 25\end{eqnarray} The first solution of the inequality \begin{eqnarray}2x+3 &\leq& 25 \\\nonumber 2x &\leq& 22/:2\\\nonumber x&\leq& 11\end{eqnarray} The second solution of the inequality \begin{eqnarray}-(2x+3) &\leq& 25 /\cdot (-1) \\\nonumber 2x+3 &\geq& -25 \\\nonumber 2x &\geq& -28/:2 \\\nonumber x &\geq& -\frac{28}{2}\end{eqnarray} The solution of the inequality can be written in the following form: $$ x \in \left[-\frac{3}{2}, 11\right]$$
• Example 33 - Solve the inequality \(\sqrt{3x-2} >2\)
  Solution - The expression under root must be greater than or equal to 0. \(3x-2 \geq 0 \Rightarrow 3x \geq 2/:3 \Rightarrow x \geq \frac{2}{3}\) \beqgin{eqnarray}\sqrt{3x-2} &>& 2/^2 \\\nonumber |3x -2| &>& 4 \end{eqnarray} The first solution of the inequality \begin{eqnarray}3x-2 &>& 4 \\\nonumber 3x&>& 6/:3 \\\nonumber x &>& 2\end{eqnarray} The second solution of the inequality \begin{eqnarray}3x-2 &<& -4 \\\nonumber 3x &<& -2 /:3 \\\nonumber x&<& -\frac{2}{3}\end{eqnarray}