Question

x²-4x+2=0

x²+6x=0

-3x²+9x=0

Answer 1

a) $x^{2}-4x+2=0$

$a=1,\;\;\;b=-4,\;\;\;c=2\\ \\ \\ \Delta=b^{2}-4ac\\ \\ \Delta=\left(-4 \right )^{2}-4\cdot \left(1 \right )\cdot \left(2 \right )\\ \\ \Delta=16-8\\ \\ \Delta=8\\ \\ \Delta=2^{2}\cdot 2\\ \\ \\ x=\dfrac{-b\pm\sqrt{\Delta}}{2a}\\ \\ x=\dfrac{-\left(-4 \right )\pm\sqrt{2^{2}\cdot 2}}{2 \cdot \left(1\right)}\\ \\ x=\dfrac{4\pm2\sqrt{2}}{2}\\ \\ x=\dfrac{2\cdot\left(1\pm\sqrt{2} \right )}{2}\\ \\ x=1\pm\sqrt{2}\\ \\ \boxed{ \begin{array}{rcl} x=1+\sqrt{2}&\text{ ou }&x=1-\sqrt{2} \end{array} }$


b) $x^{2}+6x=0$

$x\cdot \left(x+6 \right )=0\\ \\ \begin{array}{rcl} x=0&\text{ ou }&x+6=0 \end{array}\\ \\ \boxed{ \begin{array}{rcl} x=0&\text{ ou }&x=-6 \end{array} }$


c) $-3x^{2}+9x=0$

$-3x\cdot \left(x-3 \right )=0\\ \\ \begin{array}{rcl} -3x=0&\text{ ou }&x-3=0 \end{array}\\ \\ \boxed{ \begin{array}{rcl} x=0&\text{ ou }&x=3 \end{array} }$