Problems

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Solution: \begin{align*} && 0 &= z^4 +5z^3 +4z^2 -5z +1 \\ &&0 &= z^2 + z^{-2} + 5(z-z^{-1}) + 4 \\ &&&= (z-z^{-1})^2+2+5(z-z^{-1})+4 \\ &&&= w^2 + 5w + 6 \\ &&&= (w+3)(w+2) \\ \Rightarrow && 0 &= z-z^{-1}+3 \\ \Rightarrow && 0 &= z^2+3z-1 \\ \Rightarrow && z &= \frac{-3 \pm \sqrt{3^2+4}}{2} = \frac{-3 \pm \sqrt{13}}{2} \\ \Rightarrow && 0 &= z-z^{-1}+2 \\ \Rightarrow && 0 &= z^2+2z-1 \\ \Rightarrow && z &= \frac{-2 \pm \sqrt{2^2+4}}{2} = - 1 \pm \sqrt{2} \\ \end{align*} \begin{align*} &&0 &= 2z^8 -3z^7-12z^6 +12z^5 +22z^4-12z^3 -12 z^2 +3z +2 \\ && 0 &= 2(z^4+z^{-4}) - 3(z^3-z^{-3})-12(z^2+z^{-2})+12(z-z^{-1})+22 \\ &&&= 2\left ((z-z^{-1})^4+4(z^2+z^{-2})-6\right)-3 \left ((z-z^{-1})^3+3(z-z^{-1}) \right)-12 \left ((z-z^{-1})^2+2 \right)+12(z-z^{-1})+22 \\ &&&= 2(w^4+4(w^2+2)-6)-3w^3-9w-12w^2-24+12w+22 \\ &&&= 2 w^4-3w^3-4w^2+3w+2 \\ \Rightarrow && 0 &= 2(w^2+w^{-2})-3(w-w^{-1})-4 \\ &&&= 2((w-w^{-1})^2+2)-3(w-w^{-1})-4 \\ &&&= 2x^2-3x \\ &&&= x(2x-3) \\ \Rightarrow && 0 &= w -w^{-1} \\ \Rightarrow && w &= \pm 1 \\ \Rightarrow && \pm 1 &= z-z^{-1} \\ \Rightarrow && 0 &= z^2 \mp z-1 \\ \Rightarrow && z &= \frac{\pm 1 \pm \sqrt{5}}{2} \\ \Rightarrow && \frac32 &= w-w^{-1} \\ \Rightarrow && 0 &= 2w^2-3w -2 \\ &&&= (2w+1)(w-2) \\ \Rightarrow && 2 &= z-z^{-1} \\ \Rightarrow && 0 &= z^2-2z-1 \\ \Rightarrow && z &= 1 \pm \sqrt{2} \\ \Rightarrow && -\frac12 &= z-z^{-1} \\ \Rightarrow && 0 &= 2z^2+z-2 \\ \Rightarrow && z &= \frac{-1 \pm \sqrt{17}}{4} \\ \Rightarrow && z &\in \left \{ \frac{\pm 1 \pm \sqrt{5}}{2}, 1 \pm \sqrt{2}, \frac{-1 \pm \sqrt{17}}{4} \right \} \end{align*}

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Solution:

Since there are two turning points the derivative (a quadratic) has two distinct real roots. \begin{align*} && f'(x) &= 3x^2-2(p+q+r)x+(pq+qr+rp) \\ && 0 &< \Delta = 4(p+q+r)^2 - 4\cdot 3(pq+qr+rp) \\ \Rightarrow && (p+q+r)^2 &> 3(pq+qr+rp) \end{align*} If \(g^2 > 4h\) then \(p(x) = (x^2+gx+h)(x-k)\) has at least 2 real roots (possibly one repeated, and in particular it has two turning point, ie \begin{align*} && p'(x) &= (2x+g)(x-k)+(x^2+gx+h) \\ &&&= 3x^2+(2g-2k)x + (h-kg) \\ && 0 &< \Delta = 4(g-k)^2 - 4\cdot 3 (h-gk) \\ \Rightarrow && (g-k)^2 &> 3(h-gk) \end{align*} Pick \(g = h = 1\) and \(k = 1000\) then \((-999)^2 > 0 > 3(1-1000)\) so it is sufficient but not necessary.

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Solution: \begin{align*} && ( x^2 - a x + b ) ( x^2 + a x + c ) &= x^4 + (b+c-a^2)x^2 + a(b-c)x + bc \\ \Rightarrow && x^4 + p x^2 + q x + r &= x^4 + (b+c-a^2)x^2 + a(b-c)x + bc \\ \Rightarrow && p &= b+c-a^2 \tag{1}\\ && q &= a(b-c) \tag{2}\\ && r &= bc \tag{3} \end{align*} \begin{align*} (1): && p+a^2 &= b+ c \\ (2): && \frac{q}{a} &= b - c \\ \Rightarrow && b &= \frac12 (p+a^2 + \frac{q}{a}) \\ && c &= \frac12 (p+a^2 - \frac{q}{a}) \\ (3): && r &= \frac12 (p+a^2 + \frac{q}{a}) \frac12 (p+a^2 - \frac{q}{a}) \\ \Rightarrow && 4ra^2 &= (pa + a^3 + q)(pa+a^3-q) \\ &&&= (pa+a^3)^2 - q^2 \\ &&&= a^2(p+a^2)^2 -q^2 \\ &&&= a^2(p^2 + 2pa^2 + a^4) - q^2 \\ &&&= pa^2 + 2pa^4 + a^6 - q^2 \\ \end{align*} Therefore \(a^2\) is a root of \(u^3 + 2pu^2 + pu - q^2 = 4ru\), ie the given equation. When \(u = 0\), this equation is \(-q^2\), therefore the cubic is negative. But as \(u \to \infty\) the cubic tends to \(\infty\), therefore it must cross the \(x\)-axis and have a positive root. If \(p=-1, q = -6, r = 15\) then the cubic is: \(u^3 - 2u^2 + (1-60)u -36\) and so when \(u = 9\) we have \begin{align*} 9^3 - 2\cdot 9^2 -59 \cdot 9 -36 &= 9(9^2-2\cdot 9 - 29 -4) \\ &= 9(81 -18-59-4) \\ &= 0 \end{align*} so \(u = 9\) is a root Let \(y=z + 2\) \begin{align*} &&y^4 - 8 y^3 + 23 y^2 - 34 y + 39 &= (z+2)^4-8(z+2)^3 + 23(z+2)^2 - 34(z+2) + 39 \\ &&&= z^4+8z^3+24z^2+32z+16 - \\ &&&\quad -8z^3-48z^2-96z-64 \\ &&&\quad\quad +23z^2+92z+92 \\ &&&\quad\quad -34z-68 + 39 \\ &&&= z^4-z^2-6z+15 \end{align*} So conveniently this is \(p = -1, q = -6, r = 15\), so we know that \(a = 3\) is a sensible thing to true. \(b = \frac12(-1 + 9 + \frac{-6}{3}) = 3\) \(c = \frac12(-1+9-\frac{-6}{3}) = 5\) so \begin{align*} && z^4-z^2-6z+15 &= (z^2-3z+3)(z^2+3z+5) \\ &&y^4 - 8 y^3 + 23 y^2 - 34 y + 39 &= ((y-2)^2-3(y-2)+3)((y-2)^2+3(y-2)+5) \\ &&&= (y^2-4y+4-3y+6+3)(y^2-4y+4+3y-6+5) \\ &&&= (y^2-7y+13)(y^2-y+3) \end{align*}

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Solution:

1. If the roots are \(ak^{-1}, a, ak\) then we must have that \(p = a(k^{-1}+1+k)\), \(q = a^2(k^{-1}+k+1)\) and \(r = a^3\), therefore \(a = \frac{q}{p}\) (ie one of the roots is \(\frac q p\) and \(r = \left ( \frac{q}{p} \right)^3 \Rightarrow q^3 =rp^3 \Rightarrow q^3-rp^3 = 0\)
2. Suppose \(r = q^3/p^3\) then \(\left (\frac{q}{p} \right)^3 - p\left (\frac{q}{p} \right)^2+q\left (\frac{q}{p} \right) - r = \frac{--pq^2+pq^2}{p^2} =0 \), therefore \(q/p\) is a root by the factor theorem. We must also have the product of the three roots is \(q^3/p^3\) but one of the roots is \(q/p\) therefore the product of the other two roots is \(q^2/p^2\), but the condition \(ac = b^2\) is precisely the condition that \(a,b,c\) is a geometric progression.
3. If the three roots are \(a-d, a, a+d\) then \(p = 3a\), \(q = a^2-da+a^2+da+a^2-d^2 = 3a^2-d^2\), \(r = a(a^2-q^2)\), therefore \(\frac{p}{3}\left (q-\frac{2p^2}9 \right) = r\) Similarly, suppose \(\frac{p}{3}\) is a root, then the other two roots must sum to twice this and therefore they are in arithmetic progression. The condition \(\frac{p}{3}\) is a root is equivalent to: \(\frac{p^3}{27} - \frac{p^3}{9} + \frac{qp}{3} - r = 0\), ie exactly \(\frac{p}{3}\left (q-\frac{2p^2}9 \right) = r\), therefore this condition is both necessary and sufficient.

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Solution: Suppose \(z = a+ib\), where \(a,b \in \mathbb{R}\) then the modulus of \(z\), \(|z| = \sqrt{a^2+b^2}\). Noting the similarity to the Pythagorean theorem, we can say that \(|z_1 - z_2|\) is the distance between \(z_1\) and \(z_2\) in the Argand diagram. \begin{align*} |z_1 + z_2| &= |(z_1 - 0) + (0 -z_2)| \\ &\underbrace{\leq}_{\text{the direct distance is shorter than going via }0} |z_1 - 0| + |0 - z_2| \\ &= |z_1| + |-z_2| \\ &= |z_1| + |z_2| \end{align*} Claim: \(\displaystyle \vert\,z_1+\cdots+z_n\,\vert \leq \sum_{i=1}^n |z_i|\) Proof: (By Induction) Base Case: \(n = 1, 2\) have been proven. Inductive step, suppose it is true for \(n = k\), then consider \(n = k+1\), ie \begin{align*} \vert\,z_1+\cdots+z_k+z_{k+1}\,\vert &\leq \vert\,z_1+\cdots+z_k\vert + \vert z_{k+1}\,\vert \\ &\underbrace{\leq}_{\text{inductive hypothesis}} \sum_{i=1}^k |z_i| + |z_{k+1}| \\ &= \sum_{i=1}^{k+1} |z_i| \end{align*} Therefore if our hypothesis is true for \(n = k\) it is true for \(n = k+1\), and so since it is true for \(n = 1\) it is true by the principle of mathematical induction for all integers \(n \geq 1\). Suppose \(|z| \leq 1/4\), then consider: \begin{align*} \vert a_1z+a_2z^2+\cdots +a_nz^n \vert &\leq \vert a_1 z\vert + \vert a_2z^2\vert + \cdots + \vert a_n z_n\ \vert \\ &= \vert a_1\vert\vert z\vert + \vert a_2\vert\vert z^2\vert + \cdots + \vert a_n\vert\vert z^n\ \vert \\ &\leq 3\left ( |z| + |z|^2 + \cdots + |z|^n \right) \\ &\leq 3 \left ( \frac{1}{4} + \frac1{4^2} + \cdots + \frac{1}{4^n} \right) \\ &< 3 \frac{1/4}{1-1/4} \\ &= 1 \end{align*} Therefore we cannot have equality and there are no solutions.

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Solution:

1. Suppose \(pqr = qrs\), since the roots are positive, we can divide by \(qr\) to obtain \(p=s\) (a contradiction. Therefore all those terms are distinct.
2. \(4c^3 = pqr+qrs+rsp+spq\), \(d^4 = pqrs\). Applying AM-GM, we obtain: \begin{align*} && c^3 = \frac{ pqr+qrs+rsp+spq}{4} & > \sqrt[4]{p^3q^3r^3s^3} = d^{3} \\ \Rightarrow && c &> d \end{align*}
3. There must be a turning point between each root (since there are no repeated roots).
4. \(f'(x) = 4x^3-12ax^2+12b^2-4c^3 = 4(x^3-3ax^2+3b^2-c^3)\). Letting the roots of this polynomial be \(\alpha, \beta, \gamma\) and again applying AM-GM, we must have: \begin{align*} && b^2 = \frac{\alpha\beta + \beta \gamma+\gamma \alpha}{3} &> \sqrt[3]{\alpha^2\beta^2\gamma^2} = c^2 \\ \Rightarrow && b &> c \end{align*}
5. Again, since there are turning points between the roots of \(f'(x)\) we must have distinct roots for \(f''(x)\), ie: \(f''(x) = 3x^2-6ax+6b^2 = 3(x^2-2ax+b^2)\) has distinct real roots. But for this to occur we must have that \((2a)^2-4b^2 = 4(a^2-b^2) > 0\), ie \(a>b\)

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Solution:

1. \begin{align*} (x+y+z)^3 &= x^3+y^3+z^3+ \\ &\quad 3xy^2 + 3xz^2 + 3yx^2 + \cdots + 3zy^2 \\ &\quad\quad + 6xyz \\ (x+y+z)(xy+yz+zx) &= x^2y+x^2z + \cdots + z^2 x + 3xyz \\ x^3+y^3+z^3 &= (x+y+z)^3 - 3(xy^2 + \cdots + zy^2) - 6xyz \\ &= \alpha^3 - 3(\alpha \beta - 3\gamma)-6\gamma \\ &= \alpha^3-3\alpha \beta+3\gamma \end{align*} Since \(4 = 1^3-3\cdot1\cdot(-1) + 3 \gamma \Rightarrow \gamma = 0\), therefore one of \(x,y,z = 0\). WLOG \(x = 0\), so \(y+z = 1, y^2 + z^2 = 3 \Rightarrow y^2 + (1-y)^2 = 3 \Rightarrow y^2 -y -1 = 0 \Rightarrow y = \frac{1 \pm \sqrt{5}}{2}\), so we have \((x,y,z) = (0, \frac{1 +\sqrt{5}}{2}, \frac{1 - \sqrt{5}}{2})\) and permutations.
2. \begin{align*} A^2 &= s(s-a)(s-b)(s-c) \\ \end{align*} Notice the second part is the same as plugging \(s= 16/2 = 8\) into our polynomial Therefore \begin{align*} A^2 &= 8 \cdot (8^3 - 16 \cdot 8^2 + 81 \cdot 8 - 128) \\ &= 8 \cdot 8 (8^2 - 16 \cdot 8 + 81- 16) \\ &= 64 (-64+81-16) \\ &= 64 \end{align*} Therefore \(A = 8\)

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Solution: \(p = a+b+c, q = ab+bc+ca, r = abc\)

1. Suppose two roots are equal to each other, this means that one of the roots is also a root of the derivative. ie \begin{align*} && 0 &= x^3+qx - r \\ && 0 &= 3x^2+q \end{align*} have a common root, but this root must satisfy \(x^2 = -\frac{q}{3}\). Then \begin{align*} &&0 &= x^3 + qx - r \\ &&&= x^3 -3x^3 - r \\ &&&= -2x^3 -r \\ \Rightarrow && r^2 &= 4x^6 \\ &&&= 4 \left ( -\frac{q}{3}\right)^3 \\ \Rightarrow && 0 &= 27r^2+4q^3 \end{align*}
2. Consider \(x = z + \frac{p}{3}\), then the equation is: \begin{align*} x^{3}-px^{2}+qx-r &= (z + \frac{p}{3})^3 - p(z + \frac{p}{3})^2 + q(z + \frac{p}{3}) - r \\ &= z^3 + pz^2 + \frac{p^2}{3}z + \frac{p^3}{27} - \\ &\quad -pz^2-\frac{2p^2}{3}z-\frac{p^3}{9} + \\ &\quad\quad qz + \frac{pq}{3} - r \\ &= z^3+\left (\frac{p^2}{3}-\frac{2p^2}{3}+q \right)z + \left (\frac{p^3}{27}-\frac{p^3}{9}+\frac{pq}{3}-r \right) \\ &= z^3+\left (-\frac{p^2}{3}+q \right)z + \left (-\frac{2p^3}{27}+\frac{pq}{3}-r \right) \\ \end{align*} Since this equation must also have repeated roots we must have: \begin{align*} 4\left (-\frac{p^2}{3}+q \right)^3 + 27 \left (-\frac{2p^3}{27}+\frac{pq}{3}-r \right)^2 = 0 \end{align*} which is exactly our desired result

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Solution: Claim: \(S_p = q^p\) for \(1 \leq p \leq n-1\) Proof: When \(p = 1\), \(S_p = \Sigma_1 = q\) as expected. Note that \(\Sigma_i = 0\) for \(i = 2, \cdots, n-1\). Using \(S_p = S_{p-1}\Sigma_{1}-S_{p-2}\Sigma_{2}+\cdots+(-1)^{p-1+1}S_{1}\Sigma_{p-1}+(-1)^{p+1}p\Sigma_{p}\), we can see that \(S_p = qS_{p-q}\) when \(1 \leq p \leq n-1\), ie \(S_p = q^p\). Note that \begin{align*} S_n &= \sum \alpha_i^n \\ &= q\sum \alpha_i^{n-1} - \sum r \\ &= qS_{n-1} - nr \\ &= q^n - nr \\ \\ S_{n+1} &= \sum \alpha_i^{n+1} \\ &= q \sum \alpha_i^{n} - r \sum \alpha_i \\ &= q^{n+1} - rq \\ \\ S_{n+2} &= \sum \alpha_i^{n+2} \\ &= q \sum \alpha_i^{n+1} - r \sum \alpha_i^2 \\ &= q^{n+2} - rq^2 \\ \end{align*} Claim: \(S_{n+m} = q^{n+m} - rq^{m}\) Proof: The obvious

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Solution: Writing \(c = \cos \theta, s = \sin \theta\) then de Moivre states that: \begin{align*} && \cos 9 \theta + i \sin 9 \theta &= (c +i s)^9 \\ &&&= c^9 + 9ic^8s - 36c^7s^2-84ic^6s^3+126c^5s^4 + 126ic^4s^5 -84c^3s^6 -36ic^2s^7+9cs^8+is^9 \\ &&&= (c^9-36c^7s^2+126c^5s^3-84c^3s^6+8cs^8)+i(9c^8s-75c^6s^3+126c^4s^5-36c^2s^7+s^9) \\ \Rightarrow && \tan 9\theta &= \frac{(9c^8s-75c^6s^3+126c^4s^5-36s^2c^7+s^9)}{(c^9-36c^7s^2+126c^5s^4-84c^3s^6+8cs^8)} \\ &&&= \frac{9t-75t^3+126s^5-36t^7+t^9}{1-36t^2+126t^4-84t^6+8t^8} \\ &&&= \frac{t(t^{2}-3)(t^{6}-33t^{4}+27t^{2}-3)}{(3t^{2}-1)(3t^{6}-27t^{4}+33t^{2}-1)} \end{align*} If we consider \(\tan 9\theta = 0\) it will have the roots \(\theta = \frac{n \pi}{9}, n \in \mathbb{Z}\), in particular, the numerator of our fraction for \(\tan 9 \theta\) will be zero for \(t = 0, \tan \frac{\pi}{9}, \tan \frac{2\pi}{9}, \tan \frac{3\pi}{9}, \tan \frac{4 \pi}{9}, \tan \frac{5\pi}{9}, \tan \frac{6 \pi}{9}, \tan \frac{7 \pi}{9}, \tan \frac{8\pi}{9}\). It's worth noting all other values of \(\theta\) will repeat these values. Also note that \(0,\tan \frac{\pi}{3}, \tan \frac{2\pi}{3}\) are the roots of \(t\) and \(t^2-3\) respectively. Therefore the other values are the roots of our sextic. However, also note that \(\tan \frac{8\pi}{9} = - \tan \frac{\pi}{9}\) and similar, therefore we can notice that all the roots in pairs can be mapped to \(\tan \frac{\pi}{9}, \tan \frac{2 \pi}{9}\) and \(\tan \frac{4 \pi}{9}\) and all those values are squared, so the roots of: \(x^3 - 33x^2+27x-3\) will be \(\tan^2 \frac{\pi}{9}, \tan^2 \frac{2 \pi}{9}\) and \(\tan^2 \frac{4 \pi}{9}\). The product of the roots will be \(3\), so \begin{align*} && \tan^2 \frac{\pi}{9} \tan^2 \frac{2 \pi}{9} \tan^2 \frac{4 \pi}{9} &= 3 \\ \Rightarrow && \tan \frac{\pi}{9} \tan \frac{2 \pi}{9} \tan \frac{4 \pi}{9} &= \pm \sqrt{3} \\ \underbrace{\Rightarrow}_{\text{all positive}} && \tan \frac{\pi}{9} \tan \frac{2 \pi}{9} \tan \frac{4 \pi}{9} &= \sqrt{3} \\ \end{align*} Notice that \(x^3 + y^3 +z^3 - 3xyz = (x+y+z)((x+y+z)^2-3(xy+yz+zx))\) Therefore \begin{align*} \tan^{6}\left(\frac{\pi}{9}\right)+\tan^{6}\left(\frac{2\pi}{9}\right)+\tan^{6}\left(\frac{4\pi}{9}\right) &= 33(33^2-3\cdot27) + 3 \cdot 3 \\ &= 33\,273 \end{align*}

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Solution: Given that \(e^{i \theta} = \cos \theta + i \sin \theta\) we must have that

1. \begin{align*} \cos 4 \theta &= \textrm{Re} \l e^{i 4 \theta} \r \\ &= \textrm{Re} \l (\cos \theta + i \sin \theta)^4 \r \\ &= \cos^4 \theta - \binom{4}{2}\cos^2 \theta \sin^2 \theta +\sin^4 \theta \\ &= \cos^4 \theta - 6\cos^2 \theta (1-\cos^2 \theta) +(1-\cos^2 \theta)^2 \\ &= 8\cos^4 \theta - 8\cos^2 \theta + 1 \end{align*}
2. Similarly, \begin{align*} \cos 6 \theta &= \textrm{Re} \l e^{i 6 \theta} \r \\ &= \textrm{Re} \l (\cos \theta + i \sin \theta)^6 \r \\ &= \cos^6 \theta -\binom{6}{2}\cos^4 \theta \sin^2 \theta +\binom{6}{4} \cos^2\theta \sin^4 \theta - \sin^6 \theta \\ &= \cos^6 \theta - 15 \cos^4 \theta (1-\cos^2 \theta) + 15\cos^2 \theta (1-\cos^2\theta)^2 - (1-\cos^2 \theta)^3\\ &= 31\cos^6 \theta-45\cos^4\theta+15\cos^2\theta-1+3\cos^2 \theta-3\cos^4 \theta+\cos^6 \theta \\ &= 32 \cos^6 \theta-48\cos^4 \theta+18\cos^2 \theta-1 \end{align*}

\begin{align*} 0 &= 16x^{6}-28x^{4}+13x^{2}-1\\ &= \frac12 (32x^6-56x^4+26x^2-1) \\ &= \frac12(32x^6-48x^4+18x^2-1-(8x^4-8x^2+1)) \end{align*} Therefore if \(x = \cos \theta\) then we are looking at solving \(\cos 6 \theta = \cos 4 \theta\). \(\cos 6 \theta - \cos 4 \theta = -2 \sin 5\theta \sin \theta = 0\). So we should be looking at \(\sin 5 \theta = 0\) and \(\sin \theta = 0\). \(\sin \theta = 0 \Rightarrow x = \cos \theta = \pm 1\) both of which are roots. The other roots will be \(\cos \frac{\pi}{5}, \cos \frac{2\pi}{5}\) etc but it's unclear this is an acceptable form. Alternatively, given our two roots, we can factorize \begin{align*} 0 &= 16x^{6}-28x^{4}+13x^{2}-1 \\ &= (x^2-1)(16x^4-12x^2+1) \end{align*} We can solve \(16y^2-12y+1=0\) to see that \(x^2 = \frac{3 \pm \sqrt{5}}{8}\) so our roots are: \(x = -1, 1, \pm \sqrt{\frac{3 + \sqrt{5}}{8}}, \pm \sqrt{\frac{3 -\sqrt{5}}{8}}\) (We might notice that \(3+\sqrt{5} =\l \frac{1+\sqrt{5}}{\sqrt{2}} \r^2\) so our final answer could be: \(x = -1, 1, \pm \frac{1+\sqrt{5}}{4}, \pm \frac{\sqrt{5}-1}{4}\))