Question

Пожалуйста нужно срочно! С объяснением кое каким хотяб

Answer 1

$\displaystyle\bf\\4)\\\\x^{2} -6x-2=0\\\\D=(-6)^{2} -4\cdot(-2)=36+8=44=(2\sqrt{11} ) ^{2} \\\\\\x_{1} =\frac{6-2\sqrt{11} }{2} =3-\sqrt{11} \\\\\\x_{2} =\frac{6+2\sqrt{11} }{2} =3+\sqrt{11}$

Корни нового уравнения на 3 больше , значит они равны :

$\displaystyle\bf\\x_{1} =3-\sqrt{11} +3=6-\sqrt{11} \\\\x_{2} =3+\sqrt{11} +3=6+\sqrt{11} \\\\Teorema \ Vieta:\\\\x^{2} +px+q=0\\\\x_{1} +x_{2} =-p\\\\x_{1} \cdot x_{2} =q\\\\x_{1} +x_{2} =6-\sqrt{11} +6+\sqrt{11} =12\\\\x_{1} \cdot x_{2} =(6-\sqrt{11} )\cdot(6+\sqrt{11} )=6^{2} -(\sqrt{11} )^{2} =36-11=25\\\\\\Otvet \ : \ x^{2} -12x+25=0$

$\displaystyle\bf\\\\5)\\\\x^{2} -(a-3)x+(a^{2} +a)=0\\\\x_{1} \cdot x_{2} =a^{2} +a \ -teorema \ Vieta\\\\a^{2} +a=2\\\\a^{2} +a-2=0\\\\D=1^{2} -4\cdot(-2)=1+8=9=3^{2} \\\\\\a_{1} =\frac{-1-3}{2} =-2\\\\\\a_{2} =\frac{-1+3}{2} =1$

Ответ : при  a = - 2  и  a = 1