Precalculus: Mathematics for Calculus, 7th Edition

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