Question

Solve the quadratic equation numerically (using tables of x- and y-values).
x^2 +
+ 2x + 1 = 0
x = -1 or x = -1
x= -3 or x = -3
b.I x= -4 or x= -3
d. x = 2 or x = -1
a
C.
Please select the best answer from the choices provided
A
B
ОО

Answer 1

The best answer from the choices provide is x = -1 or x = -1

Given the expression

$x^2 + 2x + 1 = 0$

Factorizing the given expression

$(x^2 +x)+(x + 1) = 0$

Factor out the common variable from both brackets

$x(x+1)+1(x+1)=0\\(x+1)(x+1)=0\\x+1 \ and \ x+1=0\\x =0-1 \ and \ x=0-1\\x=-1 \ and \ x =-1$

Hence the best option is x = -1 or x = -1

Learn more about factorization here: https://brainly.com/question/16099992

Answer 2

The solutions for $f(x) = x^{2}+2\cdot x + 1 = 0$ are $x_{1} = x_{2} = -1$.

We can estimate the roots by a numerical approach, which consists in evaluating the quadratic function ($f(x) = x^{2}+2\cdot x + 1 = 0$) for a set of values of $x$. We consider the set of values between $-4\le x \le 4$:

$\,\,\,\,\,\,x \,\,\,\,\,\,\,\,f(x)\\-4\,\,\,\,\,\,\,\,\,\,\,\,9\\-3\,\,\,\,\,\,\,\,\,\,\,\,4\\-2\,\,\,\,\,\,\,\,\,\,\,\,1\\-1\,\,\,\,\,\,\,\,\,\,\,\,0\\.\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,1\\.\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,4\\.\,\,\,2\,\,\,\,\,\,\,\,\,\,\,\,\,9\\.\,\,\,3\,\,\,\,\,\,\,\,\,\,\,\,\,16\\.\,\,\,4\,\,\,\,\,\,\,\,\,\,\,\,\,25$

According to the previous input, we conclude that roots of the quadratic formula are $x_{1} = x_{2} = -1$.

The solutions for $f(x) = x^{2}+2\cdot x + 1 = 0$ are $x_{1} = x_{2} = -1$.