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To find the set [tex]\( A_1 \)[/tex] defined as [tex]\( A_1=\left\{x: x^2+7x+12=0, x \in \mathbb{Z} \right\} \)[/tex], we need to solve the quadratic equation [tex]\( x^2 + 7x + 12 = 0 \)[/tex] for integer values of [tex]\( x \)[/tex].
Let's solve this step-by-step:
1. Write down the quadratic equation:
[tex]\[ x^2 + 7x + 12 = 0 \][/tex]
2. Factor the quadratic equation:
We need to find two numbers that multiply to [tex]\( 12 \)[/tex] (the constant term) and add up to [tex]\( 7 \)[/tex] (the coefficient of the [tex]\( x \)[/tex]-term). These numbers are [tex]\( 3 \)[/tex] and [tex]\( 4 \)[/tex] because:
[tex]\[ 3 \cdot 4 = 12 \quad \text{and} \quad 3 + 4 = 7 \][/tex]
Hence, we can factor the quadratic equation as:
[tex]\[ (x + 3)(x + 4) = 0 \][/tex]
3. Solve the factored equation by setting each factor equal to zero:
[tex]\[ (x + 3) = 0 \quad \text{or} \quad (x + 4) = 0 \][/tex]
Solving these, we get:
[tex]\[ x + 3 = 0 \implies x = -3 \][/tex]
[tex]\[ x + 4 = 0 \implies x = -4 \][/tex]
4. Collect the integer solutions:
The integer solutions to the equation [tex]\( x^2 + 7x + 12 = 0 \)[/tex] are [tex]\( x = -3 \)[/tex] and [tex]\( x = -4 \)[/tex].
5. Form the set [tex]\( A_1 \)[/tex]:
[tex]\[ A_1 = \left\{ x : x^2 + 7x + 12 = 0, x \in \mathbb{Z} \right\} = \{ -3, -4 \} \][/tex]
Therefore, the set [tex]\( A_1 \)[/tex] is:
[tex]\[ A_1 = \{ -3, -4 \} \][/tex]
Let's solve this step-by-step:
1. Write down the quadratic equation:
[tex]\[ x^2 + 7x + 12 = 0 \][/tex]
2. Factor the quadratic equation:
We need to find two numbers that multiply to [tex]\( 12 \)[/tex] (the constant term) and add up to [tex]\( 7 \)[/tex] (the coefficient of the [tex]\( x \)[/tex]-term). These numbers are [tex]\( 3 \)[/tex] and [tex]\( 4 \)[/tex] because:
[tex]\[ 3 \cdot 4 = 12 \quad \text{and} \quad 3 + 4 = 7 \][/tex]
Hence, we can factor the quadratic equation as:
[tex]\[ (x + 3)(x + 4) = 0 \][/tex]
3. Solve the factored equation by setting each factor equal to zero:
[tex]\[ (x + 3) = 0 \quad \text{or} \quad (x + 4) = 0 \][/tex]
Solving these, we get:
[tex]\[ x + 3 = 0 \implies x = -3 \][/tex]
[tex]\[ x + 4 = 0 \implies x = -4 \][/tex]
4. Collect the integer solutions:
The integer solutions to the equation [tex]\( x^2 + 7x + 12 = 0 \)[/tex] are [tex]\( x = -3 \)[/tex] and [tex]\( x = -4 \)[/tex].
5. Form the set [tex]\( A_1 \)[/tex]:
[tex]\[ A_1 = \left\{ x : x^2 + 7x + 12 = 0, x \in \mathbb{Z} \right\} = \{ -3, -4 \} \][/tex]
Therefore, the set [tex]\( A_1 \)[/tex] is:
[tex]\[ A_1 = \{ -3, -4 \} \][/tex]
Answer:
Step-by-step explanation:To find the elements of the set \( A_1 \), we need to solve the quadratic equation \( x^2 + 7x + 12 = 0 \) for \( x \) in the set of integers \( \mathbb{Z} \).
First, we solve the quadratic equation using the factorization method. We look for two numbers that multiply to \( 12 \) (the constant term) and add up to \( 7 \) (the coefficient of \( x \)).
The quadratic equation is:
\[ x^2 + 7x + 12 = 0 \]
We can factorize this equation as:
\[ (x + 3)(x + 4) = 0 \]
Now, we solve for \( x \) by setting each factor equal to zero:
\[ x + 3 = 0 \quad \text{or} \quad x + 4 = 0 \]
\[ x = -3 \quad \text{or} \quad x = -4 \]
Therefore, the elements of the set \( A_1 \) are \( -3 \) and \( -4 \).
Thus,
\[ A_1 = \{ -3, -4 \} \]
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