Answer
The solution set for the quadratic equation, $-{{x}^{2}}-2x+1=0$ is $\left\{ -1\pm \sqrt{2} \right\}$.
Work Step by Step
Consider the quadratic equation,
$-{{x}^{2}}-2x+1=0$
Now, compare the equation with the standard quadratic equation $a{{x}^{2}}+bx+c=0\text{ , }\left( a\ne 0 \right)$.
Here, $a=-1,\text{ }b=-2\text{ and }c=1$
Recall that the quadratic formula is given by,
$x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
Substitute $-1$ for a, $-2$ for b and 1 for c.
$x=\frac{-\left( -2 \right)\pm \sqrt{{{\left( -2 \right)}^{2}}-4\left( -1 \right)\left( 1 \right)}}{2\left( -1 \right)}$
Simplify the radical.
$\begin{align}
& x=\frac{-\left( -2 \right)\pm \sqrt{{{\left( -2 \right)}^{2}}-4\left( -1 \right)\left( 1 \right)}}{2\left( -1 \right)} \\
& =\frac{2\pm \sqrt{4+4}}{-2} \\
& =\frac{2\pm \sqrt{8}}{-2}
\end{align}$
Take the square root.
$\begin{align}
& x=\frac{2\pm \sqrt{4+4}}{-2} \\
& =\frac{2\pm \sqrt{8}}{-2} \\
& =\frac{2\pm 2\sqrt{2}}{-2}
\end{align}$
Strike off the common factor.
$\begin{align}
& x=\frac{2\pm 2\sqrt{2}}{-2} \\
& =\frac{2\left( 1\pm \sqrt{2} \right)}{-2} \\
& =-1\pm \sqrt{2}
\end{align}$
The above formula represents two values of x, one with a positive sign and one with a negative sign.
First, take the positive sign.
$x=-1+\sqrt{2}$
Secondly, take the negative sign,
$x=-1-\sqrt{2}$
Thus, the two values of x are $-1+\sqrt{2}$ and $-1\pm \sqrt{2}$.
Therefore, the solution set for the quadratic equation $-{{x}^{2}}-2x+1=0$ is $\left\{ -1\pm \sqrt{2} \right\}$.
Verify whether the solution set $\left\{ -1\pm \sqrt{2} \right\}$ satisfies the equation $-{{x}^{2}}-2x+1=0$.
Substitute the value of x in the left hand side of the equation and check if it’s equal to the right hand side.
Substitute $-1+\sqrt{2}$ for x,
$\begin{align}
& -{{x}^{2}}-2x+1=-{{\left( -1+\sqrt{2} \right)}^{2}}-2\left( -1+\sqrt{2} \right)+1 \\
& =-1-2+2\sqrt{2}+2-2\sqrt{2}+1 \\
& =0
\end{align}$
Thus, the right hand side of the equation is equal to the left hand side of the equation.
Substitute $-1-\sqrt{2}$ for x,
$\begin{align}
& -{{x}^{2}}-2x+1=-{{\left( -1-\sqrt{2} \right)}^{2}}-2\left( -1-\sqrt{2} \right)+1 \\
& =-1-2-2\sqrt{2}+2+2\sqrt{2}+1 \\
& =0
\end{align}$
Thus, the right hand side of the equation is equal to the left hand side of the equation.
Therefore, the solution set for the quadratic equation $-{{x}^{2}}-2x+1=0$ is $\left\{ -1\pm \sqrt{2} \right\}$.
