Question
In the following, determine whether the given values are solution of the given equation or not:
$x^2+x+1=0, x=0, x=1$
$x^2+x+1=0, x=0, x=1$
✓
Answer
We have been given that,
$x^2+x+1=0, x=0, x=1$
Now, if $x=0$ is a solution of the equation then it should satisfy the equation.
So, substituting $x=0$ in the equation we get
$ x^2+x+1 $
$ =(0)^2+(0)+1$
$ =1$
Hence $x=0$ is not a solution of the given quadratic equation.
Also, if $x=1$ is a solution of the equation it should satisfy the equation So, substituting $x=1$ in the equation, we get
$ x^2+x+1 $
$ =(1)^2+(1)+1 $
$ =3$
Hence $x=1$ is not a soluion of the quadratic equation.
Therefore, from the above results we find out that both $x=0$ and $x=1$ are not a solution of the given quadratic equation.
$x^2+x+1=0, x=0, x=1$
Now, if $x=0$ is a solution of the equation then it should satisfy the equation.
So, substituting $x=0$ in the equation we get
$ x^2+x+1 $
$ =(0)^2+(0)+1$
$ =1$
Hence $x=0$ is not a solution of the given quadratic equation.
Also, if $x=1$ is a solution of the equation it should satisfy the equation So, substituting $x=1$ in the equation, we get
$ x^2+x+1 $
$ =(1)^2+(1)+1 $
$ =3$
Hence $x=1$ is not a soluion of the quadratic equation.
Therefore, from the above results we find out that both $x=0$ and $x=1$ are not a solution of the given quadratic equation.
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