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:
$b=-\left({-2b \over 2(1)} \right)$
$b={-(-2b)\over 2}$
$b =b$
