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Sagot :
Sure, let's solve the given system of equations step-by-step:
1. The first equation is:
[tex]\[ x^2 + 3x = 4 \][/tex]
To solve for [tex]\( x \)[/tex], we first need to rearrange this into a standard quadratic form:
[tex]\[ x^2 + 3x - 4 = 0 \][/tex]
This is a quadratic equation in the form [tex]\( ax^2 + bx + c = 0 \)[/tex] where [tex]\( a = 1 \)[/tex], [tex]\( b = 3 \)[/tex], and [tex]\( c = -4 \)[/tex]. We can solve this quadratic equation using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Plugging the values [tex]\( a = 1 \)[/tex], [tex]\( b = 3 \)[/tex], and [tex]\( c = -4 \)[/tex] into the formula, we get:
[tex]\[ x = \frac{-3 \pm \sqrt{3^2 - 4(1)(-4)}}{2(1)} \][/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]
So, the solutions for [tex]\( x \)[/tex] are:
[tex]\[ x = \frac{-3 + 5}{2} = 1 \quad \text{and} \quad x = \frac{-3 - 5}{2} = -4 \][/tex]
Therefore, [tex]\( x = 1 \)[/tex] and [tex]\( x = -4 \)[/tex].
2. Substitute these solutions into the second equation to solve for [tex]\( y \)[/tex]. The second equation is:
[tex]\[ y^2 = x + 3 \][/tex]
Let's solve for [tex]\( y \)[/tex] for each value of [tex]\( x \)[/tex].
- For [tex]\( x = 1 \)[/tex]:
[tex]\[ y^2 = 1 + 3 = 4 \][/tex]
Taking the square root of both sides:
[tex]\[ y = \pm 2 \][/tex]
Hence, [tex]\( y = 2 \)[/tex] or [tex]\( y = -2 \)[/tex].
- For [tex]\( x = -4 \)[/tex]:
[tex]\[ y^2 = -4 + 3 = -1 \][/tex]
Taking the square root of both sides:
[tex]\[ y = \pm \sqrt{-1} \][/tex]
Since [tex]\( \sqrt{-1} = i \)[/tex], we get:
[tex]\[ y = \pm i \][/tex]
Hence, [tex]\( y = i \)[/tex] or [tex]\( y = -i \)[/tex].
To summarize, the solutions of the given system of equations are:
- For [tex]\( x = 1 \)[/tex]:
[tex]\[ y = 2 \quad \text{or} \quad y = -2 \][/tex]
- For [tex]\( x = -4 \)[/tex]:
[tex]\[ y = i \quad \text{or} \quad y = -i \][/tex]
Thus, the full set of solutions is:
[tex]\[ (x, y) = (1, 2), (1, -2), (-4, i), (-4, -i) \][/tex]
1. The first equation is:
[tex]\[ x^2 + 3x = 4 \][/tex]
To solve for [tex]\( x \)[/tex], we first need to rearrange this into a standard quadratic form:
[tex]\[ x^2 + 3x - 4 = 0 \][/tex]
This is a quadratic equation in the form [tex]\( ax^2 + bx + c = 0 \)[/tex] where [tex]\( a = 1 \)[/tex], [tex]\( b = 3 \)[/tex], and [tex]\( c = -4 \)[/tex]. We can solve this quadratic equation using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Plugging the values [tex]\( a = 1 \)[/tex], [tex]\( b = 3 \)[/tex], and [tex]\( c = -4 \)[/tex] into the formula, we get:
[tex]\[ x = \frac{-3 \pm \sqrt{3^2 - 4(1)(-4)}}{2(1)} \][/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]
So, the solutions for [tex]\( x \)[/tex] are:
[tex]\[ x = \frac{-3 + 5}{2} = 1 \quad \text{and} \quad x = \frac{-3 - 5}{2} = -4 \][/tex]
Therefore, [tex]\( x = 1 \)[/tex] and [tex]\( x = -4 \)[/tex].
2. Substitute these solutions into the second equation to solve for [tex]\( y \)[/tex]. The second equation is:
[tex]\[ y^2 = x + 3 \][/tex]
Let's solve for [tex]\( y \)[/tex] for each value of [tex]\( x \)[/tex].
- For [tex]\( x = 1 \)[/tex]:
[tex]\[ y^2 = 1 + 3 = 4 \][/tex]
Taking the square root of both sides:
[tex]\[ y = \pm 2 \][/tex]
Hence, [tex]\( y = 2 \)[/tex] or [tex]\( y = -2 \)[/tex].
- For [tex]\( x = -4 \)[/tex]:
[tex]\[ y^2 = -4 + 3 = -1 \][/tex]
Taking the square root of both sides:
[tex]\[ y = \pm \sqrt{-1} \][/tex]
Since [tex]\( \sqrt{-1} = i \)[/tex], we get:
[tex]\[ y = \pm i \][/tex]
Hence, [tex]\( y = i \)[/tex] or [tex]\( y = -i \)[/tex].
To summarize, the solutions of the given system of equations are:
- For [tex]\( x = 1 \)[/tex]:
[tex]\[ y = 2 \quad \text{or} \quad y = -2 \][/tex]
- For [tex]\( x = -4 \)[/tex]:
[tex]\[ y = i \quad \text{or} \quad y = -i \][/tex]
Thus, the full set of solutions is:
[tex]\[ (x, y) = (1, 2), (1, -2), (-4, i), (-4, -i) \][/tex]
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