#### Answer

a) The solutions are $r_{1}=4$ and $r_{2}=5$. The product of the solutions indeed equals $20$, the constant of the equation. And the sum of them does equal $9$, the negative of the coefficient of x.
b) The solutions to the first equation are $r_{1}=-2$ and $r_{1}=4$. The product of the solutions indeed equals $-8$, the constant of the equation. And the sum of them does equal $2$, the negative of the coefficient of $x$.
The solutions to the second equation are $r_{1}= -2-√2$ and $r_{2} = -2 +√2$. The multiplication does give $2$, the constant. And the sum gives $-4$, the negative of the coefficient of $x$.
c) From the quadratic formula, we get the two solutions. By multiplying them, the square roots will get eliminated and eventually, after simplifying, $c$ will equal the constant.
Similarly, by adding them, the square roots will get eliminated and eventually, after simplifying, $b$ will equal the negative of the coefficient of x. See work step by step.

#### Work Step by Step

a) First, we use the quadratic formula to find the solutions to the equation $x^2-9x+20$, where $ a = 1$, $b = -9$, and $c = 20$:
$x={-(-9)\pm\sqrt{(-9)^2-4(1)(20)} \over 2(1)}$
Simplifying we get:
$x={9\pm\sqrt{1} \over 2}$
Now we have both solutions:
$r_{1}={9-1 \over 2} = {8\over2}=4$
$r_{2}={9+1 \over 2}= {10\over2}=5$
Multiplying them does equal to the constant $c$: $5\cdot4=20$ and adding them does equal to the negative of $b$, the coefficient of x: $5+4=9$
b) For the first equation $x^2-2x-8$, we'll find the solutions using the quadratic formula where $a=1$, $b=-2$, and $c=-8$:
$x={-(-2)\pm\sqrt{(-2)^2-4(1)(-8)} \over 2(1)}$
Simplifying we get:
$x={2\pm\sqrt{36} \over 2}$
Now we have both solutions:
$r_{1}={2-6 \over 2} = {-4\over2}=-2$
$r_{2}={2+6 \over 2}= {8\over2}=4$
We can see the relationships between the solutions and coefficients. Multiplying them equals to the constant: $-2\cdot4=-8$ and adding them equals to the negative of the coefficient of x: $-2+4=2$
For the second equation $x^2+4x+2$, we'll find the solutions using the quadratic formula where $a=1$, $b=4$, and $c=2$:
$x={-4\pm\sqrt{4^2-4(1)(2)} \over 2(1)}$
Simplifying we get:
$x={-4\pm\sqrt{8} \over 2}$
We can further simplify by taking 4 out of the square root and taking 2 as a common multiple:
$x={-4\pm\sqrt{2\cdot4} \over 2}$
$x={-4\pm2\sqrt{2} \over 2}$
$x={2(-2\pm\sqrt{2}) \over 2}$
$x={-2\pm\sqrt{2}}$
Now we have both solutions:
$r_{1}={-2-\sqrt{2}}$
$r_{2}={-2+\sqrt{2}}$
We can see the relationships between the solutions and coefficients. Multiplying them equals to the constant $c$:
${(-2-\sqrt{2}})\cdot({-2+\sqrt{2}})={4-2\sqrt2+2\sqrt2-(\sqrt2)^2}=4-2=2$
And adding them equals to the negative of b, the coefficient of x:
${(-2-\sqrt{2}}) +({-2+\sqrt{2}})=-2-2-\sqrt2+\sqrt2=-2-2=-4$
c) From the quadratic formula we get r1 and r2:
$r_{1}={-b-\sqrt{b^2-4ac} \over 2a} $ and $r_{2}={-b+\sqrt{b^2-4ac} \over 2a}$
Now we’ll multiply them to prove that they are equal to the constant:
$c={-b-\sqrt{b^2-4ac} \over 2a} \cdot {-b+\sqrt{b^2-4ac} \over 2a}$
Because of the principle of the difference of squares $(a+b)·(a-b) =a^2-b^2$, we get:
$c={(-b)^2-\left(\sqrt{b^2-4ac}\right)^2 \over 4a}$
Simplifying further, we can see that the requirement is met:
$c={b^2-(b^2-4ac) \over 4a}$
$c={b^2-b^2+4ac \over 4a}$
$c={4ac \over 4a}$
$c=c$
Now we’ll prove that the addition of the solutions equals the negative of b, the coefficient of x:
$b=-\left({-b-\sqrt{b^2-4ac} \over 2a} + {-b+\sqrt{b^2-4ac} \over 2a}\right)$
We'll break the denominators into more parts:
$b=-\left({-b \over 2a} + {-\sqrt{b^2-4ac} \over 2a} + {-b \over 2a} + {\sqrt{b^2-4ac} \over 2a}\right)$
Since all the fractions have a common denominator, the square roots can cancel out, leaving us with:
$b=-\left({-b \over 2a} + {-b \over 2a}\right)$
Further simplifying and because we want to prove when a = 1, the requirment is met:
$b=-\left({-2b \over 2(1)} \right)$
$b={-(-2b)\over 2}$
$b =b$