Question

СРОЧНО 2х^2+16=0 4х^2-4х+1=0 2х^2+х=0 5х^2-4х-1=0 Х^2+7х+12=0

Answer 1

Ответ см. во вложении

Answer 2

I can help you solve these equations. Here are the steps and solutions:

- To solve $2x^2+16=0$, we can subtract 16 from both sides and then divide by 2 to get $x^2=-8$. Then we can take the square root of both sides and use the imaginary unit $i=\sqrt{-1}$ to get $x=\pm 2i\sqrt{2}$. - To solve $4x^2-4x+1=0$, we can use the quadratic formula $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$, where $a=4$, $b=-4$, and $c=1$. Plugging in these values, we get $x=\frac{4\pm\sqrt{16-16}}{8}=\frac{4\pm 0}{8}=\frac{1}{2}$. So the only solution is $x=\frac{1}{2}$. - To solve $2x^2+x=0$, we can factor out $x$ from both sides and get $x(2x+1)=0$. Then we can use the zero product property and set each factor equal to zero. We get $x=0$ or $2x+1=0$. Solving for $x$, we get $x=0$ or $x=-\frac{1}{2}$. - To solve $5x^2-4x-1=0$, we can use the quadratic formula again, where $a=5$, $b=-4$, and $c=-1$. Plugging in these values, we get $x=\frac{4\pm\sqrt{16+20}}{10}=\frac{4\pm\sqrt{36}}{10}=\frac{4\pm 6}{10}$. Simplifying, we get $x=\frac{1}{10}$ or $x=-1$. - To solve $x^2+7x+12=0$, we can factor the left side into $(x+3)(x+4)=0$. Then we can use the zero product property and set each factor equal to zero. We get $x+3=0$ or $x+4=0$. Solving for $x$, we get $x=-3$ or $x=-4$.

If you need more help with solving equations, you can use this online calculator or this tutorial.

: https://ru.symbolab.com/solver/equation-calculator : https://skysmart.ru/articles/mathematic/pokazatelnye-uravneniya