+514 \(x^2-3x+q=0 \quad \quad \quad x_{1}=5\)
\(3x^2+35x+c=0 \quad \quad \quad x_{1}={1 \over 3}\)
\(6x^2+fx-2=0 \quad \quad \quad x_{1}={1 \over 2}\)
S183-20
+14778 Bestimme den Parameter und die zweite Lösung x₂, wenn eine Lösung gegeben ist.
1. \(x^2-3x+q=0 \quad \quad \quad x_{1}=5\)
2. \(3x^2+35x+c=0 \quad \quad x_{1}={1 \over 3}\)
3. \(6x^2+fx-2=0 \quad \quad \quad x_{1}={1 \over 2}\)
Hallo Mathefan!
1.
\(\color{blue}x_1=5\\ \color{BrickRed}x^2-3x+q=0\\ 25-15+q=0\\ \color{blue}q=-10\)
\(\color{blue}x^2-3x-10=0\\ x=1,5\pm\sqrt{2,25+10}=1,5\pm3,5\\ \color{blue}x_2=-2\)
+14778 2.
\(\color{blue}x_1=\frac{1}{3}\)
\(\color{BrickRed}3x^2+35x+c=0\\ \frac{1}{3}+\frac{35}{3}+c=0\\ \color{blue}c=-12\)
\(\color{blue}3x^2+35x-12=0\\ \)
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\\ x = {-35 \pm \sqrt{35^2+4\cdot 3\cdot 12} \over 2\cdot 3}\\ x=\frac{-35\pm\sqrt{1369}}{6}=\frac{-35\pm37}{6}\\ \color{blue} x_2=-12\)
!
+14778 3.
\(\color{BrickRed}6x^2+fx-2=0 \quad \quad \quad x_{1}={1 \over 2}\\\)
\(6x^2+fx-2=0\\ 1,5+\frac{f}{2}-2=0\\ f=0,5\cdot 2\\ \color{blue}f=1\)
\(\color{blue}x_1=\frac{1}{2}\\ 6x^2+{\color{blue}1}\cdot x-2=0\\\)
\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\\ x = {-1 \pm \sqrt{1+48} \over 12}\\ x=\frac{-1\pm 7}{12}\\ \color{blue}x_2=-\frac{2}{3}\)
!
