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Sagot :
Sure, let's solve the system of equations step by step.
We are given two equations:
[tex]\[ y = 3x^2 + 6x + 4 \][/tex]
[tex]\[ y = -3x^2 + 4 \][/tex]
To find the solution, we will set the two equations equal to each other because they both equal [tex]\(y\)[/tex].
[tex]\[ 3x^2 + 6x + 4 = -3x^2 + 4 \][/tex]
First, we move all terms to one side to set up the equation for solving [tex]\(x\)[/tex]:
[tex]\[ 3x^2 + 6x + 4 + 3x^2 - 4 = 0 \][/tex]
Combine like terms:
[tex]\[ 6x^2 + 6x = 0 \][/tex]
Factor out the common term [tex]\(6x\)[/tex]:
[tex]\[ 6x(x + 1) = 0 \][/tex]
This gives us two possible solutions for [tex]\(x\)[/tex]:
[tex]\[ 6x = 0 \quad \text{or} \quad (x + 1) = 0 \][/tex]
Solving these equations, we get:
[tex]\[ x = 0 \quad \text{and} \quad x = -1 \][/tex]
Now, we need to find the corresponding [tex]\(y\)[/tex]-values for each [tex]\(x\)[/tex] by substituting these [tex]\(x\)[/tex]-values into either of the original equations. Let's use [tex]\(y = 3x^2 + 6x + 4\)[/tex].
1. For [tex]\( x = 0 \)[/tex]:
[tex]\[ y = 3(0)^2 + 6(0) + 4 = 4 \][/tex]
So, one solution is [tex]\((0, 4)\)[/tex].
2. For [tex]\( x = -1 \)[/tex]:
[tex]\[ y = 3(-1)^2 + 6(-1) + 4 = 3 - 6 + 4 = 1 \][/tex]
So, another solution is [tex]\((-1, 1)\)[/tex].
Thus, the solutions to the system of equations are:
[tex]\[ \boxed{(0, 4) \ \text{and} \ (-1, 1)} \][/tex]
We are given two equations:
[tex]\[ y = 3x^2 + 6x + 4 \][/tex]
[tex]\[ y = -3x^2 + 4 \][/tex]
To find the solution, we will set the two equations equal to each other because they both equal [tex]\(y\)[/tex].
[tex]\[ 3x^2 + 6x + 4 = -3x^2 + 4 \][/tex]
First, we move all terms to one side to set up the equation for solving [tex]\(x\)[/tex]:
[tex]\[ 3x^2 + 6x + 4 + 3x^2 - 4 = 0 \][/tex]
Combine like terms:
[tex]\[ 6x^2 + 6x = 0 \][/tex]
Factor out the common term [tex]\(6x\)[/tex]:
[tex]\[ 6x(x + 1) = 0 \][/tex]
This gives us two possible solutions for [tex]\(x\)[/tex]:
[tex]\[ 6x = 0 \quad \text{or} \quad (x + 1) = 0 \][/tex]
Solving these equations, we get:
[tex]\[ x = 0 \quad \text{and} \quad x = -1 \][/tex]
Now, we need to find the corresponding [tex]\(y\)[/tex]-values for each [tex]\(x\)[/tex] by substituting these [tex]\(x\)[/tex]-values into either of the original equations. Let's use [tex]\(y = 3x^2 + 6x + 4\)[/tex].
1. For [tex]\( x = 0 \)[/tex]:
[tex]\[ y = 3(0)^2 + 6(0) + 4 = 4 \][/tex]
So, one solution is [tex]\((0, 4)\)[/tex].
2. For [tex]\( x = -1 \)[/tex]:
[tex]\[ y = 3(-1)^2 + 6(-1) + 4 = 3 - 6 + 4 = 1 \][/tex]
So, another solution is [tex]\((-1, 1)\)[/tex].
Thus, the solutions to the system of equations are:
[tex]\[ \boxed{(0, 4) \ \text{and} \ (-1, 1)} \][/tex]
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