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Which system of equations can be graphed to find the solution(s) to [tex]\(x^2 = 2x + 3\)[/tex]?

A. [tex]\(\left\{\begin{array}{l}y = x^2 + 2x + 3 \\ y = 2x + 3\end{array}\right.\)[/tex]
B. [tex]\(\left\{\begin{array}{l}y = x^2 - 3 \\ y = 2x + 3\end{array}\right.\)[/tex]
C. [tex]\(\left\{\begin{array}{l}y = x^2 - 2x - 3 \\ y = 2x + 3\end{array}\right.\)[/tex]
D. [tex]\(\left\{\begin{array}{l}y = x^2 \\ y = 2x + 3\end{array}\right.\)[/tex]


Sagot :

To determine which system of equations can be graphed to find the solution(s) to the equation [tex]\( x^2 = 2x + 3 \)[/tex], let's start by rewriting the given equation in a way that can help us identify the system of equations.

Given:
[tex]\[ x^2 = 2x + 3 \][/tex]

First, isolate one side of the equation to form a standard quadratic equation:
[tex]\[ x^2 - 2x - 3 = 0 \][/tex]

Now let’s consider the graphing approach. One way to graphically solve for [tex]\( x \)[/tex] is to express the original equation as two functions and see where they intersect.

Rewrite the equation in terms of two separate functions [tex]\( y \)[/tex]:
1. [tex]\( f_1(x) = x^2 \)[/tex]
2. [tex]\( f_2(x) = 2x + 3 \)[/tex]

So the system of equations becomes:
[tex]\[ \begin{cases} y = x^2 \\ y = 2x + 3 \end{cases} \][/tex]

Thus, to find the solution(s) to the equation [tex]\( x^2 = 2x + 3 \)[/tex], we can graph the following system of equations and find the points where these graphs intersect:
[tex]\[ \begin{cases} y = x^2 \\ y = 2x + 3 \end{cases} \][/tex]

This matches the last option provided:
[tex]\[ \begin{cases} y = x^2 \\ y = 2x + 3 \end{cases} \][/tex]

Therefore, the system of equations that can be graphed to find the solution(s) to [tex]\( x^2 = 2x + 3 \)[/tex] is:
[tex]\[ \begin{cases} y = x^2 \\ y = 2x + 3 \end{cases} \][/tex]