Problems

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Solution:

1. Case \(b > 0\). Then \(bx+a\) is increasing and the greatest value is \(10b+a\), and the least value \(a-10b\) Case \(b=0\), then \(a\) is constant and the greatest and least value is \(a\) Case \(b < 0\), then \(bx+a\) is decreasing and the greatest value is \(-10b+a\) and the least value is \(10b+a\)
2. If \(c = 0\) we have the same cases as above. If \( c > 0\) the consider \(2cx+b\). if \(b-20c > 0\) then our function is increasing on our interval and the greatest value is \(100c+10b+a\) and the least value is \(100c-10b+a\) If \(20c+b < 0\) then our function is decreasing and that calculation is reversed. If neither of these are true, then the minimum will be when \(x = - \frac{b}{2c}\) and the max at one end point.

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Solution:

1. \(\,\) \begin{align*} && 0 &= nx^2 + 2\sqrt{3n^2+50}x + 2n + 15 \\ && 0 &> \Delta = 4(3n^2+5) - 4\cdot n \cdot (2n + 15) \\ \Leftrightarrow && 0 &> n^2-15n+5\\ \text{cv}: && n &= \frac{15 \pm \sqrt{225 - 20}}{2} \\ &&&\approx \frac{15\pm14.x}{2}\\ \Leftrightarrow &&n &\in \{1, 2, \cdots, 14\} \end{align*}
2. \(\,\) \begin{align*} && 0 &> \Delta = 4(pn^2+q) - 4\cdot n \cdot (rn+s) \\ \Leftrightarrow && 0&>(p-r)n^2-sn+q \end{align*} Which is always true if \(r > p\) and \(s^2 < 4q(p-r)\)
3. \(\,\) \begin{align*} && 0 &= x^2 + 2\sqrt{p+q}x+ r+s \\ && 0 &\leq \Delta = 4(p+q) - 4(r+s) \\ && 0 &\leq 1 + s^2/8 - s \\ \text{c.v}: && s &= \frac{1 \pm \sqrt{1-4 \cdot \frac{1}{8}}}{2\cdot\frac18} \\ &&&= 4 \pm 4\sqrt{\frac12} \\ &&&= 4 \pm 2\sqrt{2} \\ \Rightarrow && s \leq 4 - 2\sqrt{2} &\text{ or } s \geq 4 + 2\sqrt{2} \end{align*}

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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*}

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Solution: \begin{align*} x^4 - 2x^3+x^2 &= x^2(x^2-2x+1) \\ &= x^2(x-1)^2 > 0 \end{align*} Since \(x\)and \(x-1\) can't both be zero, and square cannot be negative. \begin{align*} x^2 - 8x+17 &= (x-4)^2 +1 \geq 1 > 0 \end{align*} If \(x \leq 2\) then \(x^4 - 2x^3+x^2 > 0\) and \(17-8x \geq 1\) so \(x^4-2x^3+x^2-8x+17 > 0\) If \(x > 2\) then \(x^4-2x^3 = x^3(x-2) \geq 0\) and \(x^2-8x+17 > 0\) so \(x^4-2x^3+x^2-8x+17 > 0\), so for all real numbers our polynomial is positive and therefore cannot have any roots. Note that: \(x^4-x^3+x^2 = x^2(x^2-x+1) > 0\) and \(x^2-4x+4 =(x-2)^2 \geq 0\) If \(x < 1\) then \(x^4-x^3+x^2 > 0\) and \(4(1-x) > 0\) so \(x^4-x^3+x^2-4x+4 > 0\). If \(x > 1\) then \(x^4-x^3 = x^3(x-1) > 0\) and \(x^2-4x+4 \geq 0\) therefore \(x^4-x^3+x^2-4x+4 > 0\). Therefore \(x^4-x^3+x^2-4x+4 > 0\) for all real \(x\) and hence there are no real roots.

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Solution:

1. \begin{align*} \left(x-a\right)^{2}+\left(2x-b\right)^{2}+\left(3x-c\right)^{2} &= x^2-2ax+a^2 + 4x^2 -4bx+b^2 + 9x^2-6cx + c^2 \\ &= (1^2 + 2^2 + 3^2)x^2 - 2x(a+2b+3c) +a^2+b^2 + c^2 \\ &= 14x^2 - 2x(7x) + a^2 + b^2 + c^2 \\ &= a^2 + b^2 + c^2 \end{align*}
2. \begin{align*} \left(2x-a\right)^{2}+\left(3x-b\right)^{2}+\left(3x-c\right)^{2} &= (2^2+3^2+3^2)x^2 - 2x(2a+3b+3c) + (a^2 + b^2+c^2) \\ &= 22x^2 - 2x(11x) + a^2+b^2+c^2 \\ &= a^2+b^2+c^2 \end{align*}

The general result is: If \(\frac{A^2+B^2+C^2}{2}x =Aa+Bb+Cc\) then \((Ax-a)^2 + (Bx-b)^2 + (Cx-c)^2 = a^2+b^2+c^2\) Alternatively, if \(\lambda = \frac{2\mathbf{x} \cdot \mathbf{y}}{\Vert x \Vert^2}\) then \(\Vert \lambda \mathbf{x} - \mathbf{y}\Vert^2 = \Vert \mathbf{y} \Vert^2\) which is easy to see is true.