Functions, Equations and Polynomials
Solutions

1. Subtract the first equation from the second, rearrange the resulting expression and then factor to obtain $$\begin{array}{rl}-8x+y+xy-8& =0\\ xy-8x+y-8& =0\\ x(y-8)+y-8& =0\\ (x+1)(y-8)& =0\end{array}$$ Therefore, $x=-1$ or $y=8$. If $x=-1$, then substituting into the first equation and solving we obtain that $y=-9$. If $y=8$, then substituting into the first equation and solving we obtain $x=4\pm 2\sqrt{2}$. Therefore, the solutions are $(-1,-9)$ and $(4\pm 2\sqrt{2},8)$.

2. Solution 1

   We are asked for the $x$ value of the midpoint of zeros, which is the $x$ value of the vertex of the parabola. The equation is written in vertex form already and so $a=1$.

   Solution 2

   Find the $x$-intercepts: $$\begin{array}{rl}(x-1{)}^2-4& =0\\ (x-1{)}^2& =4\\ x& =1\pm 2\end{array}$$ Thus, $x=3$ or $x=-1$. Thus, $a={\displaystyle \frac{-1+3}{2}}=1$.

3. 1. Consider $a=0$ and $a=1$ and find the intersection point of the resulting equations, $y=x^2$ and $y=x^2+2x+1$. Subtracting the equations we obtain $0=2x+1$. Therefore, $x=-{\displaystyle \frac{1}{2}}$ and so the intersection point is $(-{\displaystyle \frac{1}{2}},{\displaystyle \frac{1}{4}})$. Now substitute $x=-{\displaystyle \frac{1}{2}}$ into the general equation. Therefore, $$\begin{array}{rl}y& =x^2+2ax+a\\ & ={\displaystyle \frac{1}{4}}+2a\cdot \left({\displaystyle \frac{-1}{2}}\right)+a\\ & ={\displaystyle \frac{1}{4}}\end{array}$$ Since $(-{\displaystyle \frac{1}{2}},{\displaystyle \frac{1}{4}})$ satisfies the general equation, it is a point on all of the parabolas.

   2. Now $y=x^2+2ax+a=(x+a{)}^2+a-a^2$ and so the vertex is at $(-a$, $a-a^2)$. If we represent the coordinates of the vertex by $(p$,$q)$ we have $p=-a$ and $q=a-a^2$ or $q=-p^2-p$, the required parabola. Completing the square we obtain $$q=-(p^2+p+{\displaystyle \frac{1}{4}})+{\displaystyle \frac{1}{4}}=-{(p+{\displaystyle \frac{1}{2}})}^2+{\displaystyle \frac{1}{4}}$$ and so we see that the vertex of this parabola is $(-{\displaystyle \frac{1}{2}},{\displaystyle \frac{1}{4}})$, the common point found in part (a)

4. Factoring both equations we arrive at: $$\begin{array}{}\text{(1)}& p(1+r+r^2)& =26\text{(2)}& p^{2}r(1+r+r^2)& =156\end{array}$$

   From equation $(1)$ we can see neither of the factors of its left-hand side are $0$. Dividing $(2)$ by $(1)$ gives $pr=6$. Substituting this relation back into $(1)$ we get $$\begin{array}{rl}{\displaystyle \frac{6}{r}}+6+6r& =26\\ 6-20r+6r^2& =0\\ 3r^2-10r+3& =0\\ (3r-1)(r-3)& =0\end{array}$$ Therefore, $r={\displaystyle \frac{1}{3}}$ or $r=3$. Hence $(p,r)=(2,3)$ or $(18,{\displaystyle \frac{1}{3}})$.

5. We assume, on the contrary, that the coefficients are in geometric sequence. Then ${\displaystyle \frac{b}{a}}={\displaystyle \frac{c}{b}}$ which implies that $b^2=ac$. But now the discriminant $b^2-4ac=-3b^2<0$, so that the roots are not real. Thus, we have a contradiction to the condition set out in the statement of the problem and our assumption is false.

6. Let $r$ and $s$ be the integer roots. The equation can be written as $$\begin{array}{rl}a(x-r)(x-s)& =a(x^2-(r+s)x+rs)\\ & =a{x}^2-a(r+s)x+ars\\ & =a{x}^2+bx+c\end{array}$$ with $b=-a(r+s)$ and $c=ars$. Since $a,b,c$ are in arithmetic sequence, we have $$\begin{array}{rl}c-b& =b-a\\ a+c-2b& =0\\ a+ars+2a(r+s)& =0\\ 1+rs+2(r+s)& =0& \text{(we can divide by }a\text{ since }a\ne 0\text{)}\\ rs+2r+2s+4& =3\\ (r+2)(s+2)& =3\end{array}$$

   Ignoring the order of the factors, we can factor 3 as a product of two integers in two ways: $3=1(3)$ or $3=(-1)(-3)$. Therefore, the two possibilities for the roots of quadratic are: (i) $-1$ and $1$ or (ii) $-3$ and $-5$.

7. Solution 1

   Multiplying out and collecting terms results in $x^4-6x^3+8x^2+2x-1=0$. We look for a factoring with integer coefficients, using the fact that the first and last coefficients are 1 and $-1$, respectively. So $$x^4-6x^3+8x^2+2x-1=(x^2+ax+1)(x^2+bx-1)$$ where $a$ and $b$ are undetermined coefficients. However, expanding and comparing coefficients gives $a+b=-6$ and $-a+b=2$ and $ab=8$. Since all three equations are satisfied by $a=-4$ and $b=-2$, we have factored the original expression as $$x^4-6x^3+8x^2+2x-1=(x^2-4x+1)(x^2-2x-1)$$ Factoring these two quadratics gives the roots $x=2\pm \sqrt{3}$ and $x=1\pm \sqrt{2}$.

   Solution 2

   We observe that the original equation is of the form $f(f(x))=x$, where $f(x)=x^2-3x+1$. Now if we can find $x$ such that $f(x)=x$, then $f(f(x))=x$. So we solve $f(x)=x^2-3x+1=x$ which gives the first factor $x^2-4x+1$ above. With polynomial division, we can then determine that $$x^4-6x^3+8x^2+2x-1=(x^2-4x+1)(x^2-2x-1)$$ and continue as in Solution 1.

8. The vertex has $x=2$ and $y=-16$ and so $A=(2,-16)$. When $y=0$ we get $0=x^2-4x-12$ which factors to give us intercepts at $-2$ and 6. The larger value is 6, and so $B=(6,0)$. Therefore, we want the line through $(2,-16)$ and $(6,0)$. Finding the slope of the line and using the second point, the equation of the line is $$y=\left({\displaystyle \frac{0+16}{6-2}}\right)(x-6)$$ which simplifies to $y=4x-24$.

9. Solution 1

   Multiplying gives $$\begin{array}{rl}{x}^2-(b+c)x+bc& =a^2-(b+c)a+bc\\ x^2-(b+c)x+a(-a+b+c)& =0\end{array}$$ The roots are $$\begin{array}{rl}x& ={\displaystyle \frac{b+c\pm \sqrt{(b+c{)}^2-4a(-a+b+c)}}{2}}\\ & ={\displaystyle \frac{b+c\pm \sqrt{(b+c{)}^2+4a^2-4a(b+c)}}{2}}\\ & ={\displaystyle \frac{b+c\pm \sqrt{(b+c-2a{)}^2}}{2}}\end{array}$$ Thus, $x=-a+b+c$ or $x=a$.

   Solution 2

   Observe that $x=a$ is one solution. Rearranging as in the first solution we get $$x^2-(b+c)x+a(-a+b+c)=0$$ Using the sum (or the product) of the roots, we determine that other root is $x=-a+b+c$.

10. Since $x=-2$ is a solution of $x^3-7x-6=0$, we know that $x+2$ is a factor of $x^3-7x-6$. Factoring (or using long division) we obtain $$\begin{array}{rl}{x}^3-7x-6& =(x+2)(x^2-2x-3)\\ & =(x+2)(x+1)(x-3)\end{array}$$ Thus, the roots are $-2$, $-1$ and $3$.

11. Let the roots be $r$ and $s$. Using the sum of the roots and the product of the roots we obtain $$\begin{array}{rl}r+s& ={\displaystyle \frac{-4(a-2)}{4}}\\ & =2-a\end{array}$$ and $$\begin{array}{rl}rs& ={\displaystyle \frac{-8a^2+14a+31}{4}}\\ & =-2a^2+{\displaystyle \frac{7}{2}}a+{\displaystyle \frac{31}{4}}\end{array}$$ Then $$\begin{array}{rl}{r}^2+s^2& =(r+s{)}^2-2rs\\ & =(2-a{)}^2-2(-2a^2+{\displaystyle \frac{7}{2}}a+{\displaystyle \frac{31}{4}})\\ & =4-4a+a^2+4a^2-7a-{\displaystyle \frac{31}{2}}\\ & =5a^2-11a-{\displaystyle \frac{23}{2}}.\end{array}$$ It appears that the minimum value should be at the vertex of the parabola $f(a)=5a^2-11a-{\displaystyle \frac{23}{2}}$, that is, at $a={\displaystyle \frac{11}{10}}$ (found by completing the square). But we have ignored the condition that the roots are real. The discriminant of the original equation is $$\begin{array}{rl}{B}^2-4AC& =[4(a-2){]}^2-4(4)(-8a^2+14a+31)\\ & =16(a^2-4a+4)+128a^2-224a-496\\ & =144a^2-288a-432\\ & =144(a^2-2a-3)\\ & =144(a-3)(a+1).\end{array}$$ Thus, we have real roots only when $a\ge 3$ or $a\le -1$. Therefore, $a={\displaystyle \frac{11}{10}}$ cannot be our final answer, since the roots are not real for this value. However $f(a)=5a^2-11a-{\displaystyle \frac{23}{2}}$ is a parabola opening up and is symmetrical about its axis of symmetry $a={\displaystyle \frac{11}{10}}$. So we move to the nearest value of $a$ to the axis of symmetry that gives real roots, which is $a=3$.

12. Let $g(2)=k$. Since $f$ and $g$ are inverse functions, we know that $f(k)=2$. We need to solve $$\begin{array}{rl}{\displaystyle \frac{3k-7}{k+1}}& =2\\ 3k-7& =2(k+1)\\ k& =9\end{array}$$ Thus, $g(2)=9$.

13. Complete the square to obtain $$\begin{array}{rl}y& =-2x^2-4ax+k\\ & =-2(x^2+2ax+a^2)+k+2a^2\\ & =-2(x+a{)}^2+k+2a^2\end{array}$$ The vertex is at $(-a,k+2a^2)$ which we know is $(-2,7)$. Therefore, solving we obtain $a=2$ and $k=-1$.

14. Using the sum and the product of the roots we have the four equations: $$\begin{array}{rl}a+b& =-c\\ ab& =d\\ c+d& =-a\\ cd& =b\end{array}$$

    Therefore, $$\begin{array}{rl}-(c+d)+cd& =-c\\ cd-d& =0\\ d(c-1)& =0\end{array}$$ But none of $a$, $b$, $c$ or $d$ are zero, so $c=1$. Then we get $d=b$. Substituting $d=b$ into $ab=d$ we get $a=1$. Then $d=b=-2$. Thus, $a+b+c+d=-2$.

15. The most common way to do this problem uses calculus. However, we make the substitution $z=x-4$. To get $y$ in terms of $z$, try $$\begin{array}{rl}y& =x^2-2x-3\\ & =(x-4{)}^2+6x-19\\ & =(x-4{)}^2+6(x-4)+5\\ & =z^2+6z+5\end{array}$$

    Therefore, the value we want to minimize is ${\displaystyle \frac{y-4}{(x-4{)}^2}}={\displaystyle \frac{z^2+6z+1}{z^2}}=1+{\displaystyle \frac{6}{z}}+{\displaystyle \frac{1}{z^2}}$. If we now let $u={\displaystyle \frac{1}{z}}$, we have the parabola $1+6u+u^2$ which opens up and has its minimum at $u=-3$ with minimum value of $-8$. Note that since $x$ can assume any real value except $4$, we know that $z$ and $u$ will assume all real values except zero. Thus, the minimum value of this expression is $-8$.

16. Solution 1

    Since the function $g$ is linear and has positive slope, it is one-to-one and so it is invertible.

    This means that $g^{-1}(g(a))=a$ for every real number $a$ and $g(g^{-1}(b))=b$ for every real number $b$.

    Therefore, $g(f(g^{-1}(g(a))))=g(f(a))$ for every real number $a$.

    This means that $$\begin{array}{rl}g(f(a))& =g(f(g^{-1}(g(a))))\\ & =2(g(a){)}^2+16g(a)+26\\ & =2(2a-4{)}^2+16(2a-4)+26\\ & =2(4a^2-16a+16)+32a-64+26\\ & =8a^2-6\end{array}$$ Furthermore, if $b=f(a)$, then $g^{-1}(g(f(a)))=g^{-1}(g(b))=b=f(a)$. Therefore,$$f(a)=g^{-1}(g(f(a)))=g^{-1}(8a^2-6)$$Since $g(x)=2x-4$, we have $y=2g^{-1}(y)-4$ and so $g^{-1}(y)={\displaystyle \frac{1}{2}}y+2$. Therefore,$$f(a)=\frac{1}{2}(8a^2-6)+2=4a^2-1$$and so $f(\pi )=4{\pi }^2-1$.

    Solution 2

    Since the function $g$ is linear and has positive slope, it is one-to-one and so it is invertible.

    To find a formula for $g^{-1}(y)$, we start with the equation $g(x)=2x-4$, convert to $y=2g^{-1}(y)-4$ and then solve for $g^{-1}(y)$ to obtain $2g^{-1}(y)=y+4$ and so $g^{-1}(y)={\displaystyle \frac{y+4}{2}}$.

    We are given that $g(f(g^{-1}(x)))=2x^2+16x+26$.

    We can apply the function $g^{-1}$ to both sides successively to obtain $$\begin{array}{rl}f(g^{-1}(x))& =g^{-1}(2x^2+16x+26)\\ f(g^{-1}(x))& ={\displaystyle \frac{(2x^2+16x+26)+4}{2}}& \text{(knowing a formula for }{g}^{-1}\text{)}\\ f(g^{-1}(x))& =x^2+8x+15\\ f\left({\displaystyle \frac{x+4}{2}}\right)& =x^2+8x+15& \text{(knowing a formula for }{g}^{-1}\text{)}\\ f\left({\displaystyle \frac{x+4}{2}}\right)& =x^2+8x+16-1\\ f\left({\displaystyle \frac{x+4}{2}}\right)& =(x+4{)}^2-1\end{array}$$ We want to determine the value of $f(\pi )$.

    Thus, we can replace ${\displaystyle \frac{x+4}{2}}$ with $\pi$, which is equivalent to replacing $x+4$ with $2\pi$.

    Thus, $f(\pi )=(2\pi {)}^2-1=4{\pi }^2-1$.