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3.5.3 Quiz: Nonlinear Systems of Equations

Question 2 of 10

Solve the system of equations:
[tex]\[
\begin{array}{l}
y = 3x \\
y = x^2 - 4
\end{array}
\][/tex]

A. [tex]\((-2, 0)\)[/tex] and [tex]\((2, 0)\)[/tex]

B. [tex]\((-2, -6)\)[/tex] and [tex]\((2, 6)\)[/tex]

C. [tex]\((-1, -3)\)[/tex] and [tex]\((4, 12)\)[/tex]

D. [tex]\((-1, 3)\)[/tex] and [tex]\((4, 5)\)[/tex]



Answer :

GinnyAnswer
To solve the given system of equations:
[tex]\[ \begin{cases} y = 3x \\ y = x^2 - 4 \end{cases} \][/tex]

we follow these steps:

1. Substitute [tex]\( y \)[/tex] from the first equation into the second equation:
[tex]\[ 3x = x^2 - 4 \][/tex]

2. Rearrange the equation to form a standard quadratic equation:
[tex]\[ x^2 - 3x - 4 = 0 \][/tex]

3. Solve the quadratic equation using the quadratic formula [tex]\( x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -3 \)[/tex], and [tex]\( c = -4 \)[/tex]:

[tex]\[ x = \frac{{-(-3) \pm \sqrt{{(-3)^2 - 4 \cdot 1 \cdot (-4)}}}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{{3 \pm \sqrt{{9 + 16}}}}{2} \][/tex]
[tex]\[ x = \frac{{3 \pm \sqrt{25}}}{2} \][/tex]
[tex]\[ x = \frac{{3 \pm 5}}{2} \][/tex]

4. Calculate [tex]\( x \)[/tex]:
[tex]\[ x = \frac{{3 + 5}}{2} = \frac{8}{2} = 4 \][/tex]
[tex]\[ x = \frac{{3 - 5}}{2} = \frac{-2}{2} = -1 \][/tex]

5. Find the corresponding [tex]\( y \)[/tex] values for each [tex]\( x \)[/tex] using [tex]\( y = 3x \)[/tex]:

For [tex]\( x = 4 \)[/tex]:
[tex]\[ y = 3 \cdot 4 = 12 \][/tex]

For [tex]\( x = -1 \)[/tex]:
[tex]\[ y = 3 \cdot (-1) = -3 \][/tex]

Thus, the solutions to the system of equations are [tex]\((-1, -3)\)[/tex] and [tex]\((4, 12)\)[/tex].

The correct answer is C. [tex]\((-1, -3)\)[/tex] and [tex]\((4, 12)\)[/tex].