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 $c=20$:
$x={−(−9)±\sqrt{(−9)^2−4(1)(20)}\over2(1)}$
Simplifying we get:
$x={9±\sqrt1\over2}$
Now we have both solutions:
$r_{1}={9−1 \over 2}={8 \over 2}=4$
$r_{2}={9+1 \over 2}={10 \over 2}=5$
Multiplying them does equal to the constant $c$: $5⋅4=20$ 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 $c=−8$:
$x={−(−2)±\sqrt{(−2)^2−4(1)(−8)}\over2(1)}$
Simplifying we get:
$x={2±\sqrt36 \over 2}$
Now we have both solutions:
$r_{1}={2−6 \over 2}={−4 \over 2}=−2$
$r_{2}={2+6 \over 2} = {8 \over 2}=4$
We can see the relationships between the solutions and coefficients. Multiplying them equals to the constant: $−2⋅4=−8$ 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 $c=2$:
$x={−4±\sqrt{4^2−4(1)(2)}\over2(1)}$
Simplifying we get:
$x={−4±\sqrt8 \over 2}$
We can further simplify by taking $4$ out of the square root and making $2$ a common multiple in the numerator:
$x={−4±\sqrt{2⋅4} \over 2}$
$x={−4±2\sqrt2 \over 2}$
$x={2(−2±\sqrt2) \over 2}$
$x=−2±\sqrt2$
Now we have both solutions:
$r_{1}=−2−\sqrt2$
$r_{2}=−2+\sqrt2$
We can see the relationships between the solutions and coefficients. Multiplying them equals to the constant $c$:
$(−2−\sqrt2)⋅(−2+\sqrt2)=4−2\sqrt2+2\sqrt2−(\sqrt2)^2=4−2=2$
And adding them equals to the negative of $b$, the coefficient of $x$:
$(−2−\sqrt2)+(−2+\sqrt2)=−2−2−\sqrt2+\sqrt2=−2−2=−4$
c) From the quadratic formula we get $r_{1}$ and $r_{2}$:
$r_{1}={−b−\sqrt{b^2−4ac} \over 2a}$ and $r_{2}={−b+\sqrt{b^2−4ac} \over 2a}$
Now we’ll multiply them to prove that they are equal to the constant:
$c={−b−\sqrt{b^2−4ac} \over 2a}⋅{−b+\sqrt{b^2−4ac} \over 2a}$
Because of the principle of the difference of squares $(a+b)⋅(a−b)=a^2−b^2$, we get:
$c={(−b)^2−(\sqrt{b^2−4ac})^2 \over 4a}$
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 equation into more fractions:
$b=− \left({{−b\over 2a}+{−\sqrt{b^2−4ac} \over 2a}+{−b \over 2a}+{\sqrt{b^2−4ac} \over 2a} }\right)$
Since all the fractions have a common denominator, the square roots can cancel out, leaving us with:
$b=−({−b \over 2a}+{−b \over 2a})$
Further simplifying and because we want to prove when $a = 1$, the requirement is met:
$b=−{−2b \over 2(1)}$
$b={−(−2b) \over 2}$
$b=b$
