Irrational Equations: Very Difficult Problems with Solutions

Solution:
Substitute [tex]t=\sqrt{x+5}\ge 0[/tex]: [tex]x^2+t=5, t^2-x=5[/tex].
Subtracting the equations: [tex](x^2-t^2)+(x+t)=0 \iff (x+t)(x-t+1)=0[/tex]
[tex]t=-x\ (x\le 0):\ x^2-x-5=0 \Rightarrow x=\dfrac{1-\sqrt{21}}{2}[/tex]
[tex]t=x+1\ (x\ge -1):\ x^2+x-4=0 \Rightarrow x=\dfrac{-1+\sqrt{17}}{2}[/tex]
The other roots of the two quadratics violate [tex]x\le 0[/tex], resp. [tex]x\ge -1[/tex].

Solution:
Domain: [tex]x\le -3[/tex] or [tex]x\gt 1[/tex]. Substitute [tex]t=\sqrt{(x+3)(x-1)}=\sqrt{x^2+2x-3}\ge 0[/tex], [tex]x^2+2x-31=t^2-28[/tex].
Case [tex]x\gt 1:\ (x-1)\sqrt{\dfrac{x+3}{x-1}}=t, t^2+3t-28=0 \Rightarrow t=4[/tex]
[tex]x^2+2x-19=0 \Rightarrow x=-1+2\sqrt{5}\ \ (x\gt 1)[/tex]
Case [tex]x\le -3:\ (x-1)\sqrt{\dfrac{x+3}{x-1}}=-t, t^2-3t-28=0 \Rightarrow t=7[/tex]
[tex]x^2+2x-52=0 \Rightarrow x=-1-\sqrt{53}\ \ (x\le -3)[/tex]

Solution:
Domain: [tex]|x|\gt 1[/tex].
For [tex]x\lt -1[/tex] the left side is negative, so [tex]x\gt 1[/tex] and both sides are positive.
Squaring: [tex]x^2+\dfrac{2x^2}{\sqrt{x^2-1}}+\dfrac{x^2}{x^2-1}=\dfrac{1225}{144}[/tex]
[tex]x^2+\dfrac{x^2}{x^2-1}=\dfrac{x^4}{x^2-1}[/tex]; Substitute [tex]u=\dfrac{x^2}{\sqrt{x^2-1}}\gt 0[/tex]:
[tex]u^2+2u=\dfrac{1225}{144} \iff (u+1)^2=\dfrac{1369}{144} \Rightarrow u=\dfrac{25}{12}[/tex]
[tex]\dfrac{x^4}{x^2-1}=\dfrac{625}{144} \iff 144x^4-625x^2+625=0 \Rightarrow x^2=\dfrac{25}{16}, x^2=\dfrac{25}{9}[/tex]
[tex]x_1=\dfrac{5}{4}, x_2=\dfrac{5}{3}[/tex]
Check: [tex]\dfrac{5}{4}+\dfrac{5}{3}=\dfrac{35}{12}\ \checkmark[/tex]

Solution:
Domain: [tex]x\ge -1[/tex]. Substitute [tex]y=\sqrt{x+1}\ge 0, 3x+3=3y^2[/tex]:
[tex]2x^2-7xy+3y^2=0 \iff (2x-y)(x-3y)=0[/tex]
[tex]y=2x\ (x\ge 0):\ 4x^2=x+1 \Rightarrow x=\dfrac{1+\sqrt{17}}{8}[/tex]
[tex]x=3y\ (x\ge 0):\ x^2=9(x+1) \Rightarrow x=\dfrac{9+3\sqrt{13}}{2}[/tex]
The negative roots are rejected.

Solution:
[tex]4x^2-9x+5=(x-1)(4x-5), 2x^2-x-1=(x-1)(2x+1), x^2-1=(x-1)(x+1)[/tex]
Domain: [tex]x\le -1, x=1[/tex] or [tex]x\ge \dfrac{5}{4}[/tex].
[tex]x=1:\ 0-0=0\ \checkmark[/tex]
[tex]x\ge \dfrac{5}{4}[/tex]. Dividing by [tex]\sqrt{x-1}\gt 0[/tex]: [tex]\sqrt{4x-5}=\sqrt{2x+1}+\sqrt{x+1} \iff x-7=2\sqrt{2x^2+3x+1}\ \ (x\ge 7)[/tex]
[tex](x-7)^2=4(2x^2+3x+1) \iff 7x^2+26x-45=0 \Rightarrow x=\dfrac{9}{7}, x=-5[/tex] — no solution [tex](x\ge 7)[/tex].
[tex]x\le -1[/tex]. Dividing by [tex]\sqrt{1-x}\gt 0[/tex]: [tex]\sqrt{5-4x}=\sqrt{-2x-1}+\sqrt{-x-1} \iff 7-x=2\sqrt{2x^2+3x+1}[/tex]
[tex]7x^2+26x-45=0 \Rightarrow x=-5\ \checkmark[/tex]
Check: [tex]\sqrt{150}-\sqrt{54}=5\sqrt{6}-3\sqrt{6}=2\sqrt{6}=\sqrt{24}\ \checkmark[/tex]

Solution:
Domain: [tex]x\in[-1, 0)\cup[1, +\infty)[/tex].
The right side is non-negative, so [tex]x\gt 0[/tex]; [tex]x=1[/tex] is not a solution, so [tex]x\gt 1[/tex].
Substitute [tex]a=\sqrt{x-\dfrac{1}{x}}, b=\sqrt{1-\dfrac{1}{x}}[/tex]: [tex]a+b=x, a^2-b^2=x-1[/tex].
[tex]a-b=\dfrac{x-1}{x}=1-\dfrac{1}{x} \Rightarrow 2a=x-\dfrac{1}{x}+1[/tex]
[tex]u=x-\dfrac{1}{x}=a^2:\ 2\sqrt{u}=u+1 \iff \left(\sqrt{u}-1\right)^2=0 \Rightarrow u=1[/tex]
[tex]x-\dfrac{1}{x}=1 \iff x^2-x-1=0 \Rightarrow x=\dfrac{1+\sqrt{5}}{2}\ \ (x\gt 1)[/tex]
Check: [tex]a=1, b=\dfrac{\sqrt{5}-1}{2}, a+b=\dfrac{1+\sqrt{5}}{2}=x\ \checkmark[/tex]

Solution:
Domain: [tex]x\ge \sqrt{\dfrac{5}{2}}[/tex].
Substitute [tex]a=\sqrt{x+1}, b=\sqrt{2x-1}, c=\sqrt{2x^2-5}[/tex]: [tex]a^2+b^2+c^2=2x^2+3x-5, 6x^2+9x-15=3(a^2+b^2+c^2)[/tex].
[tex]\sqrt{3(a^2+b^2+c^2)}=a+b+c[/tex]
[tex]3(a^2+b^2+c^2)-(a+b+c)^2=(a-b)^2+(b-c)^2+(c-a)^2\ge 0[/tex], with equality exactly when [tex]a=b=c[/tex].
[tex]x+1=2x-1=2x^2-5 \Rightarrow x=2[/tex]
Answer: [tex]x=2[/tex]

Solution:
Let [tex]f(t)=t\left(1+\sqrt{t^2+7}\right)[/tex].
The function [tex]f[/tex] is odd and strictly increasing: for [tex]t\ge 0[/tex] it is a product of a non-negative increasing factor and a positive increasing factor, and by oddness this extends to all real [tex]t[/tex].
[tex]f(2x+1)+f(x)=0 \iff f(2x+1)=-f(x)=f(-x)[/tex]
[tex]2x+1=-x \Rightarrow x=-\dfrac{1}{3}[/tex]

Solution:
Substitute [tex]y=\sqrt[3]{7x^2+9x-4}[/tex]: [tex]y=x^3-4x^2-5x+6, y^3=7x^2+9x-4[/tex].
Adding the two equations: [tex]y^3+y=x^3+3x^2+4x+2=(x+1)^3+(x+1)[/tex]
[tex]g(t)=t^3+t[/tex] is strictly increasing, so [tex]y=x+1[/tex].
[tex](x+1)^3=7x^2+9x-4 \iff x^3-4x^2-6x+5=0 \iff (x-5)(x^2+x-1)=0[/tex]
[tex]x_1=5, x_{2,3}=\dfrac{-1\pm\sqrt{5}}{2}[/tex]

Solution:
Substitute [tex]t=\sqrt{x-2}\ge 0, x=t^2+2[/tex]: [tex]at^2-6t+2a+7=0[/tex].
Each [tex]t\ge 0[/tex] gives exactly one [tex]x[/tex], so we need exactly one root [tex]t\ge 0[/tex].
[tex]a=0:\ t=\dfrac{7}{6}\ \checkmark[/tex]
[tex]a\ne 0:\ \dfrac{D}{4}=9-a(2a+7)=-(2a+9)(a-1)[/tex]
[tex]D=0:\ a=1 \Rightarrow t=3\ \checkmark, a=-\dfrac{9}{2} \Rightarrow t=-\dfrac{2}{3}[/tex] is rejected.
[tex]D\gt 0 \iff -\dfrac{9}{2}\lt a\lt 1:\ t_1t_2=\dfrac{2a+7}{a}, t_1+t_2=\dfrac{6}{a}[/tex]
Exactly one root [tex]t\ge 0[/tex] exists when [tex]t_1t_2\le 0[/tex], i.e. [tex]-\dfrac{7}{2}\le a\lt 0[/tex] (for [tex]a=-\dfrac{7}{2}[/tex] the roots are [tex]0[/tex] and [tex]-\dfrac{12}{7}[/tex]). For [tex]0\lt a\lt 1[/tex] both roots are positive, and for [tex]-\dfrac{9}{2}\lt a\lt -\dfrac{7}{2}[/tex] both are negative.
Answer: [tex]a\in\left[-\dfrac{7}{2}, 0\right]\cup\{1\}[/tex]