Problems

Solution:

1. 

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2. \(2 \textrm{Min}(x^2,2x) = 5x-3\) tells us either \(2x^2 = 5x-3 \Rightarrow 2x^2 - 5x +3 = 0 \Rightarrow (2x-3)(x-1) \Rightarrow x = 1, \frac32\) and \(0 \leq x \leq 2\) or \(4x = 5x-3 \Rightarrow x= 3\) and \(x < 0\) or \(2 > x\), therefore our solutions are \(x = 1, \frac32, 3\)
3. We have different cases based on \(x\) vs \(-2, 0, 2\), ie Case \(x \leq -2\): \begin{align*} && mx &= \mathrm{Min}(x^2, 2x) + \mathrm{Min}(x^3, 4x) \\ &&&= 2x + x^3 \end{align*} So \(2x = 2x + x^3 \Rightarrow x^3 = 0\), but \(x \leq -2\) so no solutions. or \(6x = 2x + x^3 \Rightarrow 0 = x(x^2-4) \Rightarrow x = 0, 2, -2\) so \(x = -2\). Case \(-2 < x \leq 0\): \begin{align*} && mx &= \mathrm{Min}(x^2, 2x) + \mathrm{Min}(x^3, 4x) \\ &&&= 2x + 4x \end{align*} So \(2x = 2x + 4x\) ie \(x = 0\) which is valid. Or \(6x = 2x + 4x\) ie valid for all values in \(-2 \leq x \leq 0\) Case \(0 < x \leq 2\): \begin{align*} && mx &= \mathrm{Min}(x^2, 2x) + \mathrm{Min}(x^3, 4x) \\ &&&= x^2 + x^3 \end{align*} So \(2x = x^2 + x^3 \Rightarrow 0 = x(x^2+x-2) = x(x-1)(x+2)\) so \(x = 0, 1, -2\), but the range means \(x = 0\) or \(x = 1\) Or \(6x = x^2 + x^3 \Rightarrow 0 = x(x^2+x-6) = x(x-2)(x+3)\) so \(x = 0, 2, -3\), but the range means \(x = 0\) or \(x = 2\) Case \(2 \leq x \): \begin{align*} && mx &= \mathrm{Min}(x^2, 2x) + \mathrm{Min}(x^3, 4x) \\ &&&=2x + 4x \end{align*} So \(2x = 2x + 4x \Rightarrow x = 0\) so no solutions. Or \(6x = 2x + 4x\) so a range of solutions. Therefore the final solutions for \(m = 2\) are \(x = 0, x = 1\) and for \(m = 2\) are \(x \in [-2,0] \cup [2, \infty)\)
4. \(\mathrm{Min}(x^2, x^3)\) switches when \(x = 1\), so we must consider both limits: \begin{align*} && \frac{\d y}{\d x}\vert_{x > 1} &= 4x - 5 \\ \\ && \frac{\d y}{\d x}\vert_{x < 1} &= 6x^2 - 5 \\ \end{align*} so when \(x = 1\) the sign of the derivative changes from positive to negative, hence a local maximum. The other local maxima and minima will be when \(x = \frac54\) or \(x = \pm \sqrt{5/6}\)
   

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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:

1. \begin{align*} && x^{4}+10x^{3}+26x^{2}+10x+1 &= 0 \\ \Leftrightarrow && x^2 + 10x + 26 + 10x^{-1} + x^{-2} &= 0 \\ \Leftrightarrow && (x^2 + x^{-2} + 2) + 10(x+x^{-1}) + 24 &= 0 \\ \Leftrightarrow && y^2 + 10y + 24 &= 0 \tag{\(y = x + x^{-1}\)} \\ \Leftrightarrow && (y+6)(y+4) &= 0 \\ \Leftrightarrow && \begin{cases} x+x^{-1} = -4 \\ x+x^{-1} = -6 \\ \end{cases} \\ \Leftrightarrow && \begin{cases} x^2+4x+1 = 0 \\ x^2+6x+1 = 0 \\ \end{cases} \\ \Leftrightarrow && \boxed{\begin{cases} x = -2 \pm \sqrt{3} \\ x = -3 \pm 2\sqrt{2} \\ \end{cases}} \\ \end{align*}
2. \begin{align*} && x^{4}+x^{3}-10x^{2}-4x+16=0 &= 0 \\ \Leftrightarrow && x^2 + x - 10 - 4x^{-1} + 4x^{-2} &= 0 \\ \Leftrightarrow && (x^2+4x^{-2} - 4) + (x - 4x^{-1}) - 6 &= 0 \\ \Leftrightarrow && (x^2+4x^{-2} - 4) + (x - 4x^{-1}) - 6 &= 0 \\ \Leftrightarrow && z^2 + z - 6 &= 0 \tag{\(z = x -2x^{-1}\)} \\ \Leftrightarrow && (z+3)(z-2) &= 0 \\ \Leftrightarrow && \begin{cases} x-2x^{-1} = -3 \\ x-2x^{-1} = 2 \\ \end{cases} \\ \Leftrightarrow && \begin{cases} x^2+3x-2 = 0 \\ x^2-2x-2 = 0 \\ \end{cases} \\ \Leftrightarrow && \boxed{\begin{cases} x = \frac{-3 \pm \sqrt{17}}{2} \\ x = 1 \pm \sqrt{3} \\ \end{cases}} \\ \end{align*}