Irrational Equations: Difficult Problems with Solutions

Solution:
Domain: [tex]x\ge 1[/tex].
[tex]\sqrt{x+34}-\sqrt{x+7}=\dfrac{(x+34)-(x+7)}{\sqrt{x+34}+\sqrt{x+7}}=\dfrac{27}{\sqrt{x+34}+\sqrt{x+7}}[/tex]
The left side is strictly increasing and the right side is strictly decreasing, so the equation has at most one root.
[tex]x=2:\ \sqrt{1}+\sqrt{4}=3, \sqrt{36}-\sqrt{9}=3\ \checkmark[/tex]
Answer: [tex]x=2[/tex]

Solution:
Substitute [tex]t=2x^2+x[/tex]: [tex]\sqrt{t+1}+\sqrt{4-t}=3, -1\le t\le 4[/tex].
Squaring: [tex]5+2\sqrt{(t+1)(4-t)}=9 \Rightarrow (t+1)(4-t)=4 \iff t^2-3t=0[/tex]
[tex]t=0[/tex] or [tex]t=3[/tex]
[tex]2x^2+x=0 \Rightarrow x=0, x=-\dfrac{1}{2}[/tex]
[tex]2x^2+x-3=0 \Rightarrow x=1, x=-\dfrac{3}{2}[/tex]

Solution:
Domain: [tex]\dfrac{1}{5}\le x\lt 17[/tex]. Multiplying by [tex]\sqrt{17-x}\gt 0[/tex]:
[tex]24-\sqrt{(5x-1)(17-x)}=17-x \iff \sqrt{(5x-1)(17-x)}=x+7[/tex]
Squaring: [tex](x+7\gt 0)[/tex]
[tex]-5x^2+86x-17=x^2+14x+49 \iff x^2-12x+11=0 \Rightarrow x_1=1, x_2=11[/tex]
Both values satisfy the domain.
Check: [tex]\dfrac{24}{4}-2=4=\sqrt{16}\ \checkmark, \dfrac{24}{\sqrt{6}}-\sqrt{54}=4\sqrt{6}-3\sqrt{6}=\sqrt{6}\ \checkmark[/tex]

Solution:
Substitute [tex]a=\sqrt{x^2+x+6}, b=\sqrt{x^2-x-4}[/tex]; [tex]x^2-x-4\ge 0, a\ne b[/tex].
[tex]a+b=5(a-b) \iff 2a=3b \iff 4(x^2+x+6)=9(x^2-x-4)[/tex]
[tex]5x^2-13x-60=0 \Rightarrow x_1=5, x_2=-\dfrac{12}{5}[/tex]
For both values [tex]x^2-x-4\gt 0[/tex] and [tex]a\ne b[/tex] ([tex]a=b[/tex] only for [tex]x=-5[/tex]).

Solution:
Domain: [tex]0\le x\le 97[/tex]. Substitute [tex]u=\sqrt[4]{x}\ge 0, v=\sqrt[4]{97-x}\ge 0[/tex]: [tex]u+v=5, u^4+v^4=97[/tex].
[tex]p=uv:\ u^2+v^2=25-2p, u^4+v^4=(25-2p)^2-2p^2=97[/tex]
[tex]p^2-50p+264=0 \Rightarrow p=6[/tex] or [tex]p=44[/tex]
[tex]p=44:\ z^2-5z+44=0[/tex] — no real roots.
[tex]p=6:\ z^2-5z+6=0 \Rightarrow \{u, v\}=\{2, 3\} \Rightarrow x=2^4=16[/tex] or [tex]x=3^4=81[/tex]

Solution:
[tex]x^2+4x+4=(x+2)^2, x^2-3x-10=(x+2)(x-5), x^2-10x+25=(x-5)^2[/tex]
Substitute [tex]u=\sqrt[3]{x+2}, v=\sqrt[3]{x-5}[/tex]: [tex]u^2+uv+v^2=7[/tex].
[tex]u^3-v^3=(x+2)-(x-5)=7=(u-v)(u^2+uv+v^2) \Rightarrow u-v=1[/tex]
[tex]u^2+uv+v^2=(u-v)^2+3uv=1+3uv=7 \Rightarrow uv=2[/tex]
[tex](u, v)=(2, 1) \Rightarrow x=6[/tex]; [tex](u, v)=(-1, -2) \Rightarrow x=-3[/tex]

Solution:
Domain: [tex]x\le 6[/tex]. Substitute [tex]u=\sqrt[3]{2+x}, v=\sqrt{6-x}\ge 0[/tex]: [tex]u+v=2, u^3+v^2=8[/tex].
[tex]u^3+(2-u)^2=8 \iff u^3+u^2-4u-4=0 \iff (u+1)(u^2-4)=0[/tex]
[tex]u\in\{-1, 2, -2\} \Rightarrow v=3, 0, 4\ \ (v\ge 0\ \checkmark)[/tex]
[tex]x=u^3-2:\ x=-3, x=6, x=-10[/tex]
Check: [tex]x=-10:\ \sqrt[3]{-8}+\sqrt{16}=-2+4=2\ \checkmark[/tex]

Solution:
Domain: [tex]x\gt 0, x\ne 1[/tex]. Substitute [tex]t=\sqrt[6]{x}\gt 0, t\ne 1[/tex]: [tex]\sqrt{x}=t^3, \sqrt[3]{x}=t^2[/tex].
[tex]\dfrac{1}{t^3+t^2}+\dfrac{1}{t^3-t^2}=\dfrac{2t^3}{t^4(t^2-1)}=\dfrac{2}{t(t^2-1)}=\dfrac{1}{3}[/tex]
[tex]t^3-t-6=0 \iff (t-2)(t^2+2t+3)=0 \Rightarrow t=2\ \ (t^2+2t+3\gt 0)[/tex]
[tex]x=t^6=64[/tex]
Check: [tex]\dfrac{1}{8+4}+\dfrac{1}{8-4}=\dfrac{1}{12}+\dfrac{1}{4}=\dfrac{1}{3}\ \checkmark[/tex]

Solution:
Both radicands are positive for every [tex]x[/tex] (their discriminants are negative).
Substitute [tex]a=\sqrt{2x^2+3x+5}, b=\sqrt{2x^2-2x+5}[/tex]: [tex]a-b=x, a^2-b^2=5x[/tex].
[tex]a^2-b^2=(a-b)(a+b)=x(a+b)=5x[/tex]
[tex]x=0:\ a=b=\sqrt{5}\ \checkmark[/tex]
[tex]x\ne 0:\ a+b=5 \Rightarrow a=\dfrac{5+x}{2}[/tex]
[tex]4(2x^2+3x+5)=(x+5)^2 \iff 7x^2+2x-5=0 \Rightarrow x=-1, x=\dfrac{5}{7}[/tex]
Check: [tex]x=-1:\ \sqrt{4}-\sqrt{9}=-1\ \checkmark, x=\dfrac{5}{7}:\ \dfrac{20}{7}-\dfrac{15}{7}=\dfrac{5}{7}\ \checkmark[/tex]

Solution:
Domain: [tex]x\ge 0[/tex].
[tex]x^2+2x\sqrt{x}+2x+\sqrt{x}=(x^2+2x\sqrt{x}+x)+(x+\sqrt{x})=(x+\sqrt{x})^2+(x+\sqrt{x})[/tex]
Substitute [tex]u=x+\sqrt{x}\ge 0[/tex]: [tex]u^2+u-30=0 \Rightarrow u=5[/tex]; [tex]u=-6[/tex] is rejected.
[tex]x+\sqrt{x}=5, z=\sqrt{x}\ge 0:\ z^2+z-5=0 \Rightarrow z=\dfrac{-1+\sqrt{21}}{2}[/tex]
[tex]x=z^2=\dfrac{22-2\sqrt{21}}{4}=\dfrac{11-\sqrt{21}}{2}[/tex]

Solution:
[tex]3y^2+6y+7=3(y+1)^2+4\ge 4, 5y^2+10y+14=5(y+1)^2+9\ge 9[/tex]
So the left side is at least [tex]2+3=5[/tex], while the right side is [tex]4-2y-y^2=5-(y+1)^2\le 5[/tex]. Equality is possible only when [tex]y+1=0[/tex].
Answer: [tex]y=-1[/tex]

Solution:
Domain: [tex]x\ge -4[/tex].
Case [tex]x\ge -2:\ 3\sqrt{x+4}=1-2x, 1-2x\ge 0 \iff x\le \dfrac{1}{2}[/tex]
[tex]9(x+4)=(1-2x)^2 \iff 4x^2-13x-35=0 \Rightarrow x=5\ \left(5\gt \dfrac{1}{2}\right), x=-\dfrac{7}{4}\ \checkmark[/tex]
Case [tex]-4\le x\lt -2:\ 3\sqrt{x+4}=5+2(x+2)=9+2x\gt 0[/tex]
[tex]9(x+4)=(9+2x)^2 \iff 4x^2+27x+45=0 \Rightarrow x=-3\ \checkmark, x=-\dfrac{15}{4}\ \checkmark[/tex]

Solution:
Domain: [tex]x\le 0[/tex].
[tex]\sqrt{-x(x+1)^2}=|x+1|\sqrt{-x}[/tex]
[tex]|x+1|\sqrt{-x}+(x+1)\sqrt{-x}=0 \iff \sqrt{-x}\left(|x+1|+(x+1)\right)=0[/tex]
[tex]\sqrt{-x}=0 \iff x=0[/tex]
[tex]|x+1|=-(x+1) \iff x+1\le 0 \iff x\le -1[/tex]

Solution:
Domain: [tex]x\ge 1[/tex]. Substitute [tex]t=\sqrt{x-1}\ge 0, x=t^2+1[/tex]:
[tex]x+3-4t=t^2-4t+4=(t-2)^2, x+8-6t=t^2-6t+9=(t-3)^2[/tex]
[tex]|t-2|+|3-t|=1[/tex]
[tex]|t-2|+|3-t|\ge |(t-2)+(3-t)|=1[/tex], with equality exactly when [tex]2\le t\le 3[/tex].
[tex]2\le \sqrt{x-1}\le 3 \iff 4\le x-1\le 9 \iff 5\le x\le 10[/tex]

Solution:
Domain: [tex]x\ge \dfrac{5}{3}[/tex].
[tex]\sqrt{8x+1}-\sqrt{7x+4}=\sqrt{2x-2}-\sqrt{3x-5}[/tex]
[tex]\dfrac{x-3}{\sqrt{8x+1}+\sqrt{7x+4}}=\dfrac{3-x}{\sqrt{2x-2}+\sqrt{3x-5}}[/tex]
[tex](x-3)\left(\dfrac{1}{\sqrt{8x+1}+\sqrt{7x+4}}+\dfrac{1}{\sqrt{2x-2}+\sqrt{3x-5}}\right)=0[/tex]
The bracket is positive, so [tex]x=3[/tex].
Check: [tex]\sqrt{25}+\sqrt{4}=\sqrt{25}+\sqrt{4}\ \checkmark[/tex]

Solution:
[tex]x-a\ge 0[/tex]. Squaring:
[tex]x^2-(b+c)x+bc=x^2-2ax+a^2 \iff (b+c-2a)x=bc-a^2[/tex]
[tex]b+c-2a=(b-a)+(c-a)\gt 0 \Rightarrow x=\dfrac{bc-a^2}{b+c-2a}[/tex]
[tex]x-a=\dfrac{bc-a^2-ab-ac+2a^2}{b+c-2a}=\dfrac{(b-a)(c-a)}{b+c-2a}\gt 0\ \checkmark[/tex]