Question

when the sum of 1 and twice a negative number is subtracted from twice the square of the number, 0 results. Find the number.

Answer 1

Okay, here we have this:

Considering the provided information, we obtain the following equation:

`\begin{gathered} 2x^2-\mleft(1+2x\mright)=0 \\ 2x^2-1-2x=0 \end{gathered}`

Let's solve it to find the number:

`\begin{gathered} x_(1,\: 2)=(-\left(-2\right)\pm√(\left(-2\right)^2-4\cdot\:2\left(-1\right)))/(2\cdot\:2) \\ x_(1,\: 2)=(-\left(-2\right)\pm\:2√(3))/(2\cdot\:2) \\ x_1=(-\left(-2\right)+2√(3))/(2\cdot\:2),\: x_2=(-\left(-2\right)-2√(3))/(2\cdot\:2) \\ x=(1+√(3))/(2),\: x=(1-√(3))/(2) \end{gathered}`

Since the number is negative, then we are left with the second result, so finally we find that the number is:

`x=\frac{1-\sqrt[]{3}}{2}`