Irrational Equations: Problems with Solutions

Solution:
Domain:
$4+2x-x^2 \ge 0$
$x^2-2x-4 \le 0$
$D=20$
$x_{1,2}=\frac{2\pm\sqrt{20}}{2}=\frac{2\pm2\sqrt{5}}{2}=1\pm\sqrt5$

Domain: $x\in[1-\sqrt5, 1+\sqrt5]$

$\sqrt{4+2x-x^2}=x-2$
$4+2x-x^2=(x-2)^2$
$4+2x-x^2=x^2-4x+4$
$6x-2x^2=0$
$2x^2-6x=0$
$x(2x-6)=0$
$x=0 \in [1-\sqrt5, 1+\sqrt5]$
$x=3 \in [1-\sqrt5, 1+\sqrt5]$

Check:
$\sqrt{4+2\cdot 0-0^2}=0-2 \Leftrightarrow 2=-2$
$\Rightarrow x=0$ is not a solution

$\sqrt{4+2\cdot 3-3^2}=3-2 \Leftrightarrow 1=1$
$\Rightarrow x=3$ is a solution

Answer: $x=3$

Solution:
Domain:

$x^2-5 \ge 0$
$(x-\sqrt{5})(x+\sqrt5) \ge 0$
$x\in (-\infin, \sqrt5]\cup[\sqrt5,+\infin)$


$x + 1 \ge 0 \Leftrightarrow x \ge -1$
$x \in [-1, +\infin)$

Domain: $x\in [\sqrt5, +\infin)$


$\sqrt{x^2-5}=\sqrt{x+1}$
$x^2-5=x+1$
$x^2-x-6=0$
$D=25$
$x_1=-2 \notin [\sqrt5, +\infin)$
$x_2=3 \in [\sqrt5, +\infin)$

Check:
$\sqrt{3^2-5}=\sqrt{3+1} $
$2=2$

Answer: $x=3$

Solution:
Domain:
$3x+10\ge0 \Leftrightarrow 3x \ge -10 \Leftrightarrow x \ge -\frac{10}{3}$
$x+4 \ge 0 \Leftrightarrow x \ge -4$
Domain: $x\ge -\frac{10}{3}$


$\sqrt{3x+10}-\sqrt{x+4}=2$
$\sqrt{3x+10}=2+\sqrt{x+4}$
$(\sqrt{3x+10})^2=(2+\sqrt{x+4})^2$
$3x+10=4+4\sqrt{x+4}+x+4$
$2x+2=4\sqrt{x+4}$
$x+1=2\sqrt{x+4}$
$(x+1)^2=4(x+4)$
$x^2+2x+1=4x+16$
$x^2-2x-15=0$
$D=64$
$x_1=-3 \ge -\frac{10}{3}$
$x_2=5 \ge -\frac{10}{3}$


Check:
$\sqrt{3\cdot (-3)+ 10} - \sqrt{-3+4}=2 \Leftrightarrow 8=2 \Rightarrow x=-3$ is not a solution
$\sqrt{3\cdot 5+ 10} - \sqrt{5+4}=2 \Leftrightarrow 2=2 \Rightarrow x=5$ is a solution

Answer: $x=5$

Solution:
Domain:
$x+2\ge0 \Leftrightarrow x\ge-2$
$2x-3 \ge 0 \Leftrightarrow x \ge \frac32$

$\Rightarrow$ Domain: $x \ge \frac32$


$\sqrt{x+2} - \sqrt{2x-3}=1$
$\sqrt{x+2} =1 + \sqrt{2x-3}$
$(\sqrt{x+2})^2 =(1 + \sqrt{2x-3})^2$
$x+2 =1 + 2\sqrt{2x-3} + 2x-3$
$4-x=2\sqrt{2x-3}$
$(4-x)^2=(2\sqrt{2x-3})^2$
$16-8x + x^2=4(2x-3)$
$16-8x + x^2=8x-12$
$x^2-16x+28=0$
$D=144$
$x_1=\frac{16-12}{2}=2 \ge \frac32$

$x_1=\frac{16+12}{2}=14 \ge \frac32$


Check:
$\sqrt{2+2}-\sqrt{2\cdot 2 - 3} = 1 \Leftrightarrow 1=1 \Rightarrow x=2$ is a solution
$\sqrt{14+2}-\sqrt{2\cdot 14 - 3} = 1 \Leftrightarrow -1=1 \Rightarrow x=14$ is not a solution.

Answer: $x=2$

Solution:
Domain:
$x-3 \ge 0 \Leftrightarrow x \ge 3$
$2x+1 \ge 0 \Leftrightarrow x \ge -\frac12$
$3x+4 \ge 0 \Leftrightarrow x \ge -\frac43$

$\Rightarrow $ Domain: $x \in (3, +\infin)$


$\sqrt{x-3}+\sqrt{2x+1}=\sqrt{3x+4}$
$(\sqrt{x-3}+\sqrt{2x+1})^2=(\sqrt{3x+4})^2$
$x-3 +2\sqrt{x-3}\cdot\sqrt{2x+1}+2x+1=3x+4$
$2\sqrt{x-3}\cdot\sqrt{2x+1}=6$
$\sqrt{x-3}\cdot\sqrt{2x+1}=3$
$(x-3)\cdot(2x+1)=9$
$2x^2-6x+x-3=9$
$2x^2-6x+x-12=0$
$D=121$
$x_1=\frac{5-11}{4}=-\frac32 \notin (3, +\infin) $
$x_2=\frac{5+11}{4}=4 \in (3, +\infin)$


Check:
$\sqrt{4-3} + \sqrt{2\cdot 4+1}=\sqrt{3\cdot 4+4} \Rightarrow 4=4 \Rightarrow x=4$ is a solution.

Answer: $x=4$

Solution:
Domain: $2x+17 \ge 0 \Leftrightarrow 2x \ge -17 \Leftrightarrow x \ge -\frac{17}{2}$

$\sqrt{2x+17}-1=x$
$\sqrt{2x+17}=x + 1$
$(\sqrt{2x+17})^2=(x + 1)^2$
$2x+17=x^2 + 2x + 1$
$-x^2 +16=0$
$x^2 -16=0$
$(x -4)(x+4)=0$
$x=\pm4 \in $ the domain


Check:
$\sqrt{2\cdot(-4)+17}-1 = -4 \Leftrightarrow 3-1 = -4 \Leftrightarrow$
$ 2 = -4 \Rightarrow x=-4$ is not a solution

$\sqrt{2\cdot 4 +17}-1 = 4 \Leftrightarrow 5-1 = 4 \Leftrightarrow$
$ 4 = 4 \Rightarrow x=4$ is a solution

Answer: $x=4$

Solution:
Domain:
$x+3 \ge 0 \Leftrightarrow x \ge -3$
$2-x \ge 0 \Leftrightarrow -x \ge -2 \Leftrightarrow x \le 2$
$\Rightarrow x \in [-3, 2]$


$\sqrt{x+3}-\sqrt{2-x}=1$
$\sqrt{x+3}=1+\sqrt{2-x}$
$(\sqrt{x+3})^2=(1+\sqrt{2-x})^2$
$x+3=1+2\sqrt{2-x}+(2-x)$
$2x=2\sqrt{2-x}$
$x=\sqrt{2-x}$
$x^2=2-x$
$x^2+x-2=0$
$D=9$
$x_1=-2 \in [-3, 2]$
$x_2=1 \in [-3, 2]$


Check:
$\sqrt{-2+3}-\sqrt{2-(-2)}=1 \Leftrightarrow 1-2=1 \Leftrightarrow -1=1 \Rightarrow x=-2 $ is not a solution.

$\sqrt{1+3}-\sqrt{2-1}=1 \Leftrightarrow 2-1=1 \Leftrightarrow 1=1 \Rightarrow x=1 $ is a solution.

Answer: $x=1$

Solution:
Domain:
$1+4x-x^2 \ge 0$
$x^2-4x-1 \le 0$
$D=20$
$x_{1,2}=\frac{4\pm\sqrt{20}}{2}=2\pm\sqrt{5}$
$\Rightarrow $ Domain: $x\in[2-\sqrt{5}, 2+\sqrt{5}]$


$\sqrt{1+4x-x^2}=x-1$
$(\sqrt{1+4x-x^2})^2=(x-1)^2$
$1+4x-x^2=x^2-2x+1$
$-2x^2+6x=0$
$2x^2-6x=0$
$x^2-3x=0$
$x(x-3)=0$
$x_1=0 \in [2-\sqrt{5}, 2+\sqrt{5}]$
$x_1=3 \in [2-\sqrt{5}, 2+\sqrt{5}]$

Check:
$\sqrt{1+4\cdot0-0^2}=0-1 \Leftrightarrow 1=-1 \Rightarrow x=0$ is not a solution
$\sqrt{1+4\cdot3-3^2}=3-1 \Leftrightarrow 2=2 \Rightarrow x=3$ is a solution

Answer: $x=3$

Solution:
$1+x \ge 0 \Rightarrow x \ge -1$
$1-x \ge 0 \Rightarrow x \le 1$
$\Rightarrow $ Domain: $x \in [-1, 1]$


$\sqrt{1+x} + \sqrt{1-x}=1$
$\sqrt{1+x} =1 - \sqrt{1-x}$
$(\sqrt{1+x})^2 =(1 - \sqrt{1-x})^2$
$1+x =1 - 2\sqrt{1-x} + (1-x)$
$2x-1 = -2\sqrt{1-x}$
$1-2x = 2\sqrt{1-x}$
$(1-2x)^2 = (2\sqrt{1-x})^2$
$1-4x+4x^2 = 4(1-x)$
$1-4x+4x^2 = 4-4x$
$4x^2 = 3$
$x^2 = \frac34$
$x = \pm\frac{\sqrt{3}}{2} \in $ domain

Check:
$\sqrt{1+\frac{\sqrt{3}}{2}}+\sqrt{1-\frac{\sqrt{3}}{2}}=1 \Rightarrow \sqrt{1+0.87}+\sqrt{1-0.87}=1 \Rightarrow 1.37+0.36=1 \Rightarrow 1.73 \ne 1 \Rightarrow x=\frac{\sqrt{3}}{2}$ is not a solution.

$\sqrt{1+(-\frac{\sqrt{3}}{2})}+\sqrt{1-(-\frac{\sqrt{3}}{2})}=1 \Rightarrow \sqrt{1-0.87}+\sqrt{1+0.87}=1 \Rightarrow 0.36+1.37=1 \Rightarrow 1.73 \ne 1 \Rightarrow x=-\frac{\sqrt{3}}{2}$ is not a solution.

Answer: The equation has no real solutions.

Solution:
Domain: $\frac{x}{2}+2 \ge 0 \Leftrightarrow \frac{x}{2} \ge -2 \Leftrightarrow x \ge -4$


$x+\sqrt{\frac{x}{2}+2}=2$
$\sqrt{\frac{x}{2}+2}=2-x$
$\left(\sqrt{\frac{x}{2}+2}\right)^2=(2-x)^2$
$\frac{x}{2}+2=4-4x+x^2$
$x+4=8-8x+2x^2$
$-2x^2 +9x-4=0$
$2x^2 -9x+4=0$
$D=49$
$x_{1,2}=\frac{9\pm7}{2}$
$x_1=4 \in $ domain
$x_2=\frac12 \in $ domain

Check:
$4+\sqrt{\frac{4}{2}+2} = 6 \ne 2 \Rightarrow x=4$ is not a solution.
$\frac12+\sqrt{\frac{\frac12}{2}+2} = \frac12+\sqrt{\frac94} = 2 \Rightarrow x=\frac12$ is a solution.

Answer: $x=\frac{1}{2}$

Solution:
Domain: [tex]-2\le x\le 2[/tex]. Squaring:
[tex](2+x)-2\sqrt{(2+x)(2-x)}+(2-x)=1[/tex]
[tex]4-2\sqrt{4-x^2}=1 \Rightarrow \sqrt{4-x^2}=\dfrac{3}{2}[/tex]
Such an [tex]x[/tex] exists: [tex]x=\dfrac{\sqrt{7}}{2}[/tex] satisfies the given equality.

Solution:
A sum of two square roots (each of them non-negative) equals [tex]0[/tex] only when both radicands are [tex]0[/tex].
[tex]x^2+3x-4=0 \Rightarrow x_1=1, x_2=-4[/tex]
[tex]x=1:\ 1+12-11-2=0\ \checkmark[/tex]
[tex]x=-4:\ -64+192+44-2=170\ne 0[/tex]
Answer: [tex]x=1[/tex]

Solution:
Domain: [tex]x\ge 0, x\le 2, x^2-3x\ge 0[/tex], [tex]x^2-3x\ge 0 \iff x\le 0[/tex] or [tex]x\ge 3[/tex].
The only common point of these conditions is [tex]x=0[/tex].
Check: [tex]0+\sqrt{2}+0=\sqrt{2}\ \checkmark[/tex]

Solution:
Domain: [tex]5x-20\ge 0 \iff x\ge 4[/tex].
[tex]x\ge 4 \Rightarrow x^2\ge 16 \Rightarrow 6-x^2\le -10\lt 0[/tex]
The left side is non-negative, so the equation has no solution.

Solution:
[tex]\sqrt{1-x^2}=x-2[/tex]: Domain: [tex]-1\le x\le 1[/tex], where [tex]x-2\lt 0[/tex] — no solution.
[tex](x^2-4)\sqrt{-x-2}=0[/tex]: Domain: [tex]x\le -2[/tex]; [tex]x^2-4=0 \Rightarrow x=\pm 2[/tex], [tex]\sqrt{-x-2}=0 \Rightarrow x=-2[/tex]; exactly one solution: [tex]x=-2[/tex].
[tex]\sqrt{x-2}+\sqrt{x}=0[/tex]: Domain: [tex]x\ge 2[/tex], where [tex]\sqrt{x}\gt 0[/tex] — no solution.
[tex]\dfrac{\sqrt{x-1}}{\sqrt{x^2-1}}=0[/tex]: Domain: [tex]x\gt 1[/tex]; [tex]\sqrt{x-1}=0 \Rightarrow x=1[/tex], is not in the domain — no solution.

Solution:
Domain: [tex]x^2-5x+4\ge 0 \iff x\le 1[/tex] or [tex]x\ge 4[/tex].
[tex](x-3)\sqrt{x^2-5x+4}-2(x-3)=0 \iff (x-3)\left(\sqrt{x^2-5x+4}-2\right)=0[/tex]
[tex]x=3[/tex] is not in the domain ([tex]9-15+4=-2\lt 0[/tex]).
[tex]\sqrt{x^2-5x+4}=2 \Rightarrow x^2-5x=0 \Rightarrow x_1=0, x_2=5[/tex]
Both values satisfy the domain.

Solution:
Domain: [tex]-x\ge 0 \iff x\le 0[/tex].
Substitute [tex]t=\sqrt{-x}\ge 0[/tex], [tex]-x=t^2[/tex]:
[tex]t^2-3t-10=0 \Rightarrow t_1=5, t_2=-2[/tex]; [tex]t_2=-2[/tex] is rejected.
[tex]\sqrt{-x}=5 \Rightarrow -x=25 \Rightarrow x=-25[/tex]
Check: [tex]25-3\sqrt{25}=25-15=10\ \checkmark[/tex]

Solution:
Cubing both sides (an equivalent transformation):
[tex]x^2+15=8(1+x) \iff x^2-8x+7=0 \Rightarrow x_1=1, x_2=7[/tex]
Check: [tex]\sqrt[3]{16}=2\sqrt[3]{2}\ \checkmark, \sqrt[3]{64}=4=2\sqrt[3]{8}\ \checkmark[/tex]

Solution:
[tex]\sqrt{(x-4)^2}+\sqrt{(2x-1)^2}=10 \iff |x-4|+|2x-1|=10[/tex]
Case [tex]x\ge 4:\ (x-4)+(2x-1)=10 \Rightarrow x=5\ \checkmark[/tex]
Case [tex]\dfrac{1}{2}\le x\lt 4:\ (4-x)+(2x-1)=10 \Rightarrow x=7\notin\left[\dfrac{1}{2}, 4\right)[/tex]
Case [tex]x\lt \dfrac{1}{2}:\ (4-x)+(1-2x)=10 \Rightarrow x=-\dfrac{5}{3}\ \checkmark[/tex]

Solution:
[tex](12x^2-11x)+\sqrt{12x^2-11x}-2=0[/tex]
Substitute [tex]t=\sqrt{12x^2-11x}\ge 0[/tex]: [tex]t^2+t-2=0 \Rightarrow t_1=1, t_2=-2[/tex]; [tex]t_2=-2[/tex] is rejected.
[tex]12x^2-11x=1 \iff 12x^2-11x-1=0 \iff (x-1)(12x+1)=0[/tex]
[tex]x_1=1, x_2=-\dfrac{1}{12}[/tex]

Solution:
Domain: [tex]\dfrac{x-2}{x+1}\gt 0 \iff x\lt -1[/tex] or [tex]x\gt 2[/tex].
Substitute [tex]t=\sqrt{\dfrac{x-2}{x+1}}\gt 0[/tex], [tex]\sqrt{\dfrac{x+1}{x-2}}=\dfrac{1}{t}[/tex]:
[tex]t+\dfrac{2}{t}=3 \iff t^2-3t+2=0 \Rightarrow t_1=1, t_2=2[/tex]
[tex]t=1:\ x-2=x+1[/tex] — impossible.
[tex]t=2:\ \dfrac{x-2}{x+1}=4 \Rightarrow x-2=4x+4 \Rightarrow x=-2[/tex]
Check: [tex]\sqrt{4}+2\sqrt{\dfrac{1}{4}}=2+1=3\ \checkmark[/tex]

Solution:
Domain: [tex]x\ge 0[/tex].
[tex]\dfrac{1}{\sqrt{1+x}-\sqrt{x}}=\dfrac{\sqrt{1+x}+\sqrt{x}}{(1+x)-x}=\sqrt{1+x}+\sqrt{x}[/tex]
[tex]2\left(\sqrt{1+x}+\sqrt{x}\right)=4 \Rightarrow \sqrt{1+x}+\sqrt{x}=2[/tex]
[tex]\sqrt{1+x}-\sqrt{x}=\dfrac{1}{\sqrt{1+x}+\sqrt{x}}=\dfrac{1}{2}[/tex]
[tex]2\sqrt{x}=2-\dfrac{1}{2}=\dfrac{3}{2} \Rightarrow \sqrt{x}=\dfrac{3}{4} \Rightarrow x=\dfrac{9}{16}[/tex]