Problems

Solution:

1. \(\,\) \begin{align*} && 3 &= (x-\sqrt2)^2 \\ &&&= x^2 - 2\sqrt2 x + 2 \\ \Rightarrow && 2\sqrt2 x &= x^2-1 \\ \Rightarrow && 8x^2 &= x^4 - 2x^2 + 1 \\ \Rightarrow && 0 &= x^4 - 10x^2 + 1 \end{align*} Noticing that \((\sqrt2+\sqrt3-\sqrt2)^2 = 3\) we note that \(\sqrt2 + \sqrt3\) is a root of our quartic.
2. Suppose \(x = \sqrt2 + \sqrt3 + \sqrt5\) then \begin{align*} && 0 &= (x - \sqrt5)^4 - 10(x-\sqrt5)^2 + 1 \\ &&&= x^4 - 4\sqrt5x^3 + 30x^2-20\sqrt5 x +25 - 10x^2+20\sqrt5x -50 + 1\\ &&&= (x^4+20x^2- 24) - 4\sqrt5 x^3 \\ \Rightarrow && 80x^6 &= (x^4+20x^2-24)^2 \\ &&&= x^8 + 40x^6 + 352x^4 - 960x^2+576 \\ \Rightarrow && 0 &= x^8-40x^6 + 352x^4-960x^2+576 \end{align*} So take \(g(x) = x^8-40x^6 + 352x^4-960x^2+576\).
3. Notice that if \(p(t) = t^3-3t+1\) then \(p(t -\sqrt2) = 0\) for \(t = a,b,c\) so \begin{align*} && 0 &= (t - \sqrt2)^3 -3(t - \sqrt2) + 1 \\ &&&= t^3-3\sqrt2 t^2 + 6t - 2\sqrt2 - 3t + 3\sqrt 2 + 1 \\ &&&= (t^3+3t+1) - \sqrt2 (3t^2+1) \\ \Rightarrow && 2(3t^2+1)^2 &= (t^3+3t+1)^2 \\ \Rightarrow && 2(9t^4+6t^2+1) &= t^6 + 6t^4+2t^3+9t^2+6t+1 \\ \Rightarrow && 0 &= t^6-12t^4+2t^3-3t^2+6t-1 \end{align*}
4. \(\,\) \begin{align*} && t &= \sqrt[3]{2} + \sqrt[3]{3} \\ \Rightarrow && t^3 &= 2 + 3\sqrt[3]{12} + 3\sqrt[3]{18} + 3 \\ &&&= 5 + 3 \sqrt[3]{6}(\sqrt[3]{2} + \sqrt[3]{3}) \\ &&&= 5 + 3\sqrt[3]{6}t \\ \Rightarrow && 162t^3 &= (t^3-5)^3 \\ &&&= t^9-15t^6+75t^3 -125 \\ \Rightarrow && 0 &= t^9-15t^6-87t^3-125 \end{align*} so \(k(x) = x^9 - 15x^6-87x^3-125\)

Solution:

1. \((1+\sqrt{2})^2 = 3 + 2\sqrt{2}\) so \(\sqrt{3 + 2\sqrt{2}} = 1 + \sqrt{2}\)
2. We can always factorise any quartic in the form \((x^2+ax+b)(x^2+cx+d)\), since \(x^3\) has a coefficient of \(a+b\) we must have \(a = -b\), ie the form in the question. \begin{align*} && 0 &= (x^2+sx+p)(x^2-sx+q) \\ &&&= x^4+(p+q-s^2)x^2+s(q-p)x+pq \\ \Rightarrow && pq &= -2 \\ && s(q-p) &= 12 \\ && p+q-s^2 &= -10 \\ \\ && p+q &= s^2-10 \\ && (p+q)^2 &= (s^2-10)^2 \\ && (q-p)^2 &= \frac{12}{s^2} \\ \Rightarrow && (s^2-10)^2 &= \frac{12}{s^2} + 4pq \\ && (s^2-10)^2 &= \frac{144}{s^2} -8 \\ && 0 &= s^2(s^2-10)^2+8s^2-144 \\ &&&= s^6-20s^4+108s^2-144 \\ &&&= (s^2-2)(s^2-6)(s^2-12) \end{align*} Suppose \(s = \sqrt{2}\), and we have \begin{align*} && q-p &= 6\sqrt{2} \\ && p+q &= -8 \\ \Rightarrow && q &= 3\sqrt{2}-4 \\ && p &= -4-3\sqrt{2} \end{align*} Solving our quadratic equations, we have \begin{align*} && 0 &= x^2-\sqrt{2}x-4+3\sqrt{2} \\ \Rightarrow && x &= \frac{\sqrt{2}\pm \sqrt{2-4\cdot(-4+3\sqrt{2})}}{2} \\ &&&= \frac{\sqrt{2}\pm \sqrt{18-12\sqrt{2}}}{2} \\ &&&= \frac{\sqrt{2}\pm (2\sqrt{3}-\sqrt{6})}{2} \\ \\ && 0 &= x^2+\sqrt{2}x-3\sqrt{2}-4 \\ && x &= \frac{-\sqrt{2} \pm \sqrt{2-4\cdot(3\sqrt{3}-4)}}{2}\\ && &= \frac{-\sqrt{2} \pm \sqrt{18+12\sqrt{2}}}{2}\\ && &= \frac{-\sqrt{2} \pm (\sqrt{6}+2\sqrt{3})}{2}\\ \end{align*}

Solution:

1. Step 1: There are only three possibilities for the remainder of \(a\) when divided by \(3\), (\(0\), \(1\), \(2\)). \(a = 3m+r\). If \(r = 0\) we are done, if \(r = 1\) take \(k = m\), and \(r=2\) take \(k=(m+1)\) and we have \(a = 3k-1\) as required. Then \(a^2 = (3k\pm1)^2 =9k^2\pm6k+1 = 3(3k^32\pm2k)+1\) which is clearly \(1\) more than a square. Step 3: \begin{align*} && \frac{a}{b} &= \sqrt{2}+\sqrt{3} \\ \Rightarrow && \frac{a^2}{b^2} &= 5+2\sqrt{6} \\ \Rightarrow && \frac{a^2}{b^2}-5 &= 2\sqrt{6} \\ \Rightarrow && 24 &= \left ( \frac{a^2}{b^2}-5 \right)^2 \\ &&&= 25 + \frac{a^4}{b^4}-10\frac{a^2}{b^2} \\ \Rightarrow && -b^4 &= a^4-10a^2b^2 \\ \Rightarrow && a^4+b^4 &= 10a^2b^2 \end{align*} Step 4: If \(a\) is divisible by \(3\) then \(b^4 = 10a^2b^2-a^4\) is a multiple of \(3\), but if \(b\) was not a multiple of \(3\) then \(b^2\) would be \(1\) more than a multiple of \(3\) (by Step 3) and \(b^4\) would also be \(1\) more than a multiple of \(3\), and we would have a contradiction. Step 5: Follows since either \(a,b\) are both divisible by \(3\) (contradicting Step 2), or neither is, but then \(a^2\) and \(b^2\) are both one more than a multiple of \(3\) and the RHS is one more than a multiple of \(3\) but the LHS is \(2\) more than a multiple of \(3\) which is a contradiction.
2. Step 1: If \(a\) is not divisible by \(5\) then \(a^2 \equiv \pm 1 \pmod{5}\) Step 2: Suppose \(\frac{a}{b} = \sqrt{6}+\sqrt{7}\) Step 3: \begin{align*} && \frac{a}{b} &= \sqrt{6}+\sqrt{7} \\ \Rightarrow && \frac{a^2}{b^2} &= 13 + 2\sqrt{42} \\ \Rightarrow && 168 &= \left (\frac{a^2}{b^2} - 13 \right)^2 \\ &&&= 169 - 26 \frac{a^2}{b^2} + \frac{a^4}{b^4} \\ \Rightarrow && a^4+b^4 &= 26a^2b^2 \end{align*} Step 4: If \(a\) is a multiple of \(5\) then so is \(b^4\) and hence so is \(b^2\) and \(b\). Step 5: But the left hand side is always \(2 \pmod{5}\) and the right hand side is never \(2 \pmod{5}\) contradiction. Divisibility by \(3\) doesn't work here since mod \(3\) we can have \(a = 1, b = 1\) and have a valid solution.

Solution:

1. \begin{align*} && 0 &= x + 3\sqrt{x} - \frac12 \\ \sqrt{x} = y: && 0&= y^2 + 3y - \frac12 \\ \Rightarrow && y &= \frac{-3\pm\sqrt{3^2+2}}{2} \\ &&&= \frac{-3 \pm \sqrt{11}}{2} \\ y > 0: && x &= \left ( \frac{\sqrt{11}-3}{2} \right)^2 \end{align*}
2. • \begin{align*} && 0 &= x + 10\sqrt{x+2} - 22 \\ y = \sqrt{x+2}: && 0 &= y^2 - 2 + 10y - 22 \\ &&&= y^2 + 10y - 24 \\ &&&= (y-2)(y+12) \\ \Rightarrow && y &= 2, -12 \\ y > 0: && x &= 2 \end{align*}
   • Let \(y = \sqrt{2x^2-8x-3}\), so \begin{align*} && 0 &= x^2 - 4x +\sqrt{2x^2-8x-3} - 9 \\ && 0 &= \frac{y^2+3}{2} + y - 9 \\ &&&= \frac12 y^2 +y - \frac{15}{2} \\ &&&= \frac12 (y-3)(y+5) \\ \Rightarrow && y &= 3,-5 \\ y > 0: && 9 &= 2x^2-8x-3 \\ \Rightarrow && 0 &= 2x^2-8x-12 \\ &&&= 2(x^2-4x-6) \\ \Rightarrow && x &= 2 \pm \sqrt{10} \end{align*}

Solution:

1. • \(\bf (a)\) \((1, \sqrt2)\) is an integer \(2\)-rational point. \((\frac35 + \frac45\sqrt2, \frac45 - \frac{3}{5}\sqrt2)\) is a non-integer \(2\)-rational point.
   • [\bf(b)] First notice that \((\sqrt2)^2 +3^2 = 11\) so then consider \((1 + \tfrac32\sqrt2, 1-\tfrac32\sqrt2)\) will work as \(\pi/4\) degree rotation.
   • [\bf(c)] First notice \(3^2-(\sqrt2)^2 = 2\). Notice that \((k\sec \theta + \sqrt{m} \tan \theta)^2 - (k\tan \theta + \sqrt{m} \sec \theta)^2 = k^2-m\). Taking \(k= 3\) we have \((3 \cdot \frac{13}{5} + \frac{12}{5}\sqrt{2}, 3\cdot\frac{12}5+\frac{13}{5}\sqrt2)\)

Solution:

1. Claim: For all \(n \geq 1\) \(T_{2n} = A_{2n} + B_{2n}\sqrt{a(a+1)}\) where \(A_{2n}, B_{2n}\) are integers and \(A_{2n}^2 = a(a+1)B_{2n}^2+1\) Proof: (By induction) Base case: \(n =1\). \begin{align*} && T_2 &= (\sqrt{a+1}+\sqrt{a})^2 \\ &&&= a+1+2\sqrt{a(a+1)}+a \\ &&&=2a+1+2\sqrt{a(a+1)} \\ \Rightarrow && A_2 &= 2a+1 \\ && B_2 &= 2 \\ && A_2^2 &= 4a^2+4a+1 \\ && a(a+1)B_1^2 + 1 &= 4a^2+4a+1 \end{align*} Therefore our base case is true. Suppose it is true for some \(n = k\) then consider \(n = k+1\) we must have \(T_{2k} = A_{2k}+B_{2k}\sqrt{a(a+1)}\) \begin{align*} && T_{2(k+1)} &= T_{2k} (\sqrt{a+1}+\sqrt{a})^2 \\ &&&= \left (A_{2k}+B_{2k}\sqrt{a(a+1)} \right)\left (2a+1+2\sqrt{a(a+1)} \right) \\ &&&= (2a+1)A_{2k}+2a(a+1)B_{2k} + (2A_{2k}+(2a+1)B_{2k})\sqrt{a(a+1)} \\ \Rightarrow && A_{2(k+1)} &= (2a+1)A_{2k}+2a(a+1)B_{2k} \in \mathbb{Z} \\ && B_{2(k+1)} &= 2A_{2k}+(2a+1)B_{2k} \in \mathbb{Z} \\ && A_{2(k+1)}^2 &= \left ( (2a+1)A_{2k}+2a(a+1)B_{2k} \right)^2 \\ &&&= (2a+1)^2A_{2k}^2+4a^2(a+1)^2B_{2k}^2+4a(a+1)(2a+1)A_{2k}B_{2k} \\ &&&= a^2(a+1)^2(2a+1)^2B_{2k}^2+(2a+1)^2\\ &&&\quad\quad+4a^2(a+1)^2B_{2k}^2+4a(a+1)(2a+1)A_{2k}B_{2k}\\ && a(a+1)B_{2(k+1)}^2 + 1 &= a(a+1)\left ( 2A_{2k}+(2a+1)B_{2k} \right)^2 \\ &&&= 4a(a+1)A_{2k}^2 + 4a(a+1)(2a+1)A_{2k}B_{2k} + a(a+1)(2a+1)^2B_{2k}^2 + 1\\ &&&= 4a^2(a+1)^2B_{2k}^2+4a(a+1) + \\ &&&\quad\quad+ 4a(a+1)(2a+1)A_{2k}B_{2k} + a(a+1)(2a+1)^2B_{2k}^2 + 1 \end{align*} So our relation holds ad therefore by induction we're done.
2. When \(n\) is odd, notice that \begin{align*} && T_n &= (\sqrt{a+1}+\sqrt{a})(A_m+B_m\sqrt{a(a+1)})\\ &&&= \sqrt{a+1}(\underbrace{A_m+aB_m}_{C_n})+\sqrt{a}(\underbrace{(a+1)B_m+A_m}_{D_n}) \\ \\ && (a+1)C_n^2 &= (a+1)(A_m+aB_m)^2 \\ &&&= (a+1)(A_m^2+2aA_mB_m+B_m^2) \\ &&&= (a+1)(a(a+1)B_m^2+1+2aA_mB_m+B_m^2) \\ &&&= a(a+1)^2B_m^2 + 2a(a+1)A_mB_m+(a+1)B_m^2+a+1 \\ && aD_n^2+1 &= a((a+1)^2B_m + 2(a+1)A_mB_m + A_m^2)+1 \\ &&&= a(a+1)^2B_m + 2a(a+1)A_mB_m + aB_{m}^2+a+1 \end{align*}
3. For even \(n\) \(T_n = \sqrt{a(a+1)B_{2n}^2+1} + \sqrt{a(a+1)B_{2n}^2}\) For odd \(n\), \(T_n = \sqrt{aD_n^2+1}+ \sqrt{aD_n^2}\) therefore it is always the sum of the square root of two consecutive integers.

Solution: \begin{align*} && \tan \left ( \tfrac14 \pi -\tfrac12 x \right) &\equiv \frac{\tan \tfrac{\pi}{4} - \tan \tfrac12 x}{1 + \tan \tfrac{\pi}{4} \tan \tfrac12 x} \\ &&&= \frac{1-\tan \tfrac12 x}{1+\tan \tfrac12 x} \\ \\ && \sec x - \tan x &= \frac{1+t^2}{1-t^2} - \frac{2t}{1-t^2} \\ &&&= \frac{(1-t)^2}{(1-t)(1+t)} \\ &&&= \frac{1-t}{1+t} \end{align*} Therefore both sides are equal to the same thing.

1. \(\tan \tfrac18 \pi = \tan(\tfrac14 \pi - \tfrac12 \tfrac14\pi) = \sec \tfrac14 \pi - \tan \tfrac14 \pi = \sqrt{2} - 1\) \begin{align*} && \tan \tfrac{11}{24} \pi &= \tan (\tfrac13 \pi +\tfrac18 \pi) \\ &&&= \frac{\tan \tfrac13 \pi +\tan \tfrac18 \pi}{1-\tan \tfrac13 \pi \tan \tfrac18 \pi} \\ &&&= \frac{\sqrt{3} + \sqrt{2} - 1}{1 - \sqrt{3}(\sqrt{2}-1)} \\ &&&= \frac{\sqrt{3} + \sqrt{2} - 1}{1 +\sqrt{3}-\sqrt{6}} \\ \end{align*}
2. \(\,\) \begin{align*} && (\sqrt{3}-\sqrt{6}+1)(2+\sqrt{2}+\sqrt{3}+\sqrt{6}) &= (2\sqrt{3}+\sqrt{6}+3+3\sqrt{2}) + \\ &&&\quad+(-2\sqrt{6}-2\sqrt{3}-3\sqrt{2}-6) + \\ &&&\quad+(2+\sqrt{2}+\sqrt{3}+\sqrt{6}) \\ &&&= \sqrt{3}+\sqrt{2}-1 \end{align*}
3. \(\,\) \begin{align*} && \tan \tfrac{1}{48} \pi &= \tan (\tfrac14\pi - \tfrac{11}{48} \pi) \\ &&&= \sec \tfrac{11}{24} \pi - \tan \tfrac{11}{24} \pi \\ &&&= \sqrt{1+\tan^2 \tfrac{11}{24}\pi} - \tan \tfrac{11}{24} \pi \\ &&&= \sqrt{1 + (2+\sqrt{2}+\sqrt{3}+\sqrt{6})^2} - (2+\sqrt{2}+\sqrt{3}+\sqrt{6}) \\ &&&= \sqrt{16+10\sqrt{2} + 8\sqrt{3}+6\sqrt{6}} - 2 - \sqrt2 - \sqrt3-\sqrt6 \end{align*}

Solution:

1. First notice that \(\frac{1}{5+\sqrt{24}} = \frac{5-\sqrt{24}}{25-24} = 5 - \sqrt{24}\), hence \begin{align*} && ( 5 + \sqrt {24})^4 + \frac{1 }{(5 + \sqrt {24})^4} &= ( 5 + \sqrt {24})^4 + ( 5 - \sqrt {24})^4 \\ \end{align*} where clearly all terms including \(\sqrt{24}\) will cancel out, therefore it is an integer. \begin{align*} && 5 + \sqrt{24} &< 5 + 5 = 10 \\ \Rightarrow && \frac{1}{5+\sqrt{24}}& > \frac{1}{10} = 0.1 \\ && 2(5 + \sqrt{24}) &=10 + \sqrt{96} > 19 \\ \Rightarrow && \frac{1}{5+\sqrt{24}} & < \frac{2}{19} < \frac{2}{18} = \frac19 = 0.11111\ldots < 0.11 \end{align*} Therefore, \(10^{-4} < (5+\sqrt{24})^4 < 0.11^{-4} = 0.00014641\) \begin{align*} && (5+\sqrt{24})^4 + (5-\sqrt{24})^4 &= 2(5^4+6\cdot5^2\cdot24+24^2) \\ &&&= 2\cdot (625 + 3600+576) \\ &&&= 9602 \\ \Rightarrow && (5+\sqrt{24})^4 &= 9602 - \epsilon, \epsilon \in (0.0001, 0.00014641) \\ \Rightarrow && (5+\sqrt{24})^4 &\in (9601.999854, ,9601.9999) \\ \Rightarrow && (5+\sqrt{24})^4 &= 9601.9998 \, (4 \text{ d.p.}) \end{align*}
2. Notice that \((N+\sqrt{N^2-1})^{k}+(N-\sqrt{N^2-1})^{k}\) is an integer for the same reason as before (sum of conjugates). Notice also that \(\frac{1}{N+\sqrt{N^2-1}} = N - \sqrt{N^2-1}\) and that so it sufficies to show that \begin{align*} && N + \sqrt{N^2-1} &> 2N-\tfrac12 \\ \Leftrightarrow && \sqrt{N^2-1} &> N - \tfrac12 \\ \Leftrightarrow && N^2-1 &> N^2-N+1\\ \Leftrightarrow && N &> \tfrac32\\ \end{align*} Which is true since \(N > 1\) and \(N\) is an integer.

Solution:

1. \begin{align*} (3+2\sqrt{5})^3 &= 3^3 + 3 \cdot 3^2 \cdot 2\sqrt{5} + 3 \cdot 3 \cdot (2 \sqrt{5})^2 + (2\sqrt{5})^3 \\ &= 27 + 180 + (54+40)\sqrt{5} \\ &= 207 + 94\sqrt{5} \end{align*}
2. \begin{align*} && (c-d\sqrt{2})^3 &= c^3+6cd -(3c^2d+2d^3)\sqrt{2} \\ \Rightarrow && 99 &= c(c^2+6d^2) \\ && 70 &= d(3c^2+2d^2) \\ \Rightarrow && c & \mid 99, d \mid 70 \\ && c &= 3, d = 2 \end{align*}
3. \begin{align*} && 0 &= x^6 - 198x^3 + 1 \\ \Rightarrow && 0 &= (x^3-99)^2+1-99^2 \\ \Rightarrow && x^3 &= 99 \pm \sqrt{99^2-1} \\ &&&= 99 \pm 10 \sqrt{98} \\ &&&= 99 \pm 70 \sqrt{2} \\ \Rightarrow && x &= 3 \pm 2 \sqrt{2} \end{align*}

Solution:

1. \(\,\) \begin{align*} && 0 &= \sqrt{3x^2+1} + \sqrt{x} -2x-1 \\ \Rightarrow && 2x+1 &= \sqrt{3x^2+1} + \sqrt{x} \\ \Rightarrow && 4x^2+4x+1 &= 3x^2+1+x+2\sqrt{x(3x^2+1)} \\ \Rightarrow && x^2+3x &= 2\sqrt{x(3x^2+1)} \\ \Rightarrow && x^2(x^2+6x+9) &= 4x(3x^2+1) \\ \Rightarrow && 0 &= x^4-6x^3+9x^2-4x \\ &&&= x(x-4)(x-1)^2 \end{align*} So clearly we have \(x = 0, x = 1, x = 4\). \(x = 0\) works, \(x = 1\) works, \(x = 4\) works.
2. \(\,\) \begin{align*} && 0 &= \sqrt{3x^2+1} - 2\sqrt{x} +x-1 \\ \Rightarrow && 1-x &= \sqrt{3x^2+1}-2\sqrt{x} \\ \Rightarrow && x^2-2x+1 &= 3x^2+1+4x-4\sqrt{x(3x^2+1)} \\ \Rightarrow && x^2+3x &= 2\sqrt{x(3x^2+1)} \\ \Rightarrow && 0 &= x(x-4)(x-1)^2 \end{align*} Again we must check \(x = 0, x = 1, x = 4\). \(x = 0,1\) work, but \(x = 4\) is not a solution.
3. \(\,\) \begin{align*} && 0 &= \sqrt{3x^2+1} - 2\sqrt{x} -x+1 \\ \Rightarrow && x-1 &= \sqrt{3x^2+1} - 2\sqrt{x} \\ \Rightarrow && x^2-2x+1 &= 3x^2+1+4x-4\sqrt{x(3x^2+1)} \\ \Rightarrow && x^2+3x &= 2\sqrt{x(3x^2+1)}\\ \Rightarrow && 0 &= x(x-4)(x-1)^2 \end{align*} So again, we need to check \(x = 0, 1, 4\). \(x = 0, 4\) work, but \(x = 1\) fails.

Solution: \begin{align*} && f'(x) &= \frac12 ax^{-1/2}-\frac12(x-b)^{-1/2} \\ \Rightarrow f'(x) = 0: && 0 &= \frac{a\sqrt{x-b}-\sqrt{x}}{\sqrt{x(x-b)}} \\ \Rightarrow && x &= a^2(x-b)\\ \Rightarrow && x &= \frac{a^2b}{a^2-1} \\ && f(x) &= a^2 \sqrt{\frac{b}{a^2-1}} - \sqrt{\frac{a^2b}{a^2-1}-b} \\ &&&= a^2 \sqrt{\frac{b}{a^2-1}} - \sqrt{\frac{b}{a^2-1}} \\ &&&= \sqrt{b(a^2-1)} \end{align*}

+ - ↺

1. \(c^2 = 16\), \(2 \cdot (3^2-1) = 16\), so we should have exactly one solution at \(x = \frac{3^2 \cdot 2}{3^2 -1 } = \frac{9}{4}\)
2. \(c^2 = 25\) and \(3 \cdot (3^2 - 1) = 24, 3 \cdot (3^2) = 27\), so we look for two solutions. \begin{align*} && 5 & = 3 \sqrt{x} - \sqrt{x-3} \\ \Rightarrow && 25 &= 9x+x-3-6\sqrt{x(x-3)} \\ \Rightarrow && 3\sqrt{x(x-3)} &= 5x-14 \\ \Rightarrow && 9x(x-3) &= 25x^2-140x+196 \\ \Rightarrow && 0 &= 16x^2-113x+196 \\ &&&= (x-4)(16x-49) \\ \Rightarrow && x &= 4, \frac{49}{16} \end{align*}

Solution:

1. \(\,\)
   

   + - ↺
   \begin{align*} && 1+2x-x^2 = 2/x \\ \Rightarrow && 0 &= x^3-2x^2-x+2 \\ &&&= (x+1)(x^2-3x+2) \\ &&&= (x+1)(x-1)(x-2) \end{align*} Therefore the inequality is satisfied on \((1,2)\) and \((-1,0)\)
2. \(\,\)
   

   + - ↺
   \begin{align*} && \sqrt{3x+10} &= 2+\sqrt{x+4} \\ && 3x+10 &= x+8 + 4\sqrt{x+4} \\ && 16(x+4) &= 4(x+1)^2 \\ && 4x+16 &= x^2+2x+1 \\ \Rightarrow && 0 &= x^2-2x-15 \\ &&&= (x-5)(x+3) \end{align*} Therefore \(x > 5\)