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Sagot :
To solve the problem of finding two integers that add up to 3 and multiply to -70, we will follow a systematic approach:
1. Let the two integers be [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
- We need to find [tex]\( x \)[/tex] and [tex]\( y \)[/tex] such that their sum is 3 and their product is -70.
- Therefore, we have two equations:
[tex]\[ x + y = 3 \quad \text{(1)} \][/tex]
[tex]\[ x \cdot y = -70 \quad \text{(2)} \][/tex]
2. Express [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex] from equation (1):
- From equation (1), we can express [tex]\( y \)[/tex] as:
[tex]\[ y = 3 - x \][/tex]
3. Substitute [tex]\( y \)[/tex] into equation (2):
- Substitute [tex]\( y \)[/tex] from the above expression into equation (2):
[tex]\[ x \cdot (3 - x) = -70 \][/tex]
- Simplify this equation to get a quadratic equation:
[tex]\[ 3x - x^2 = -70 \][/tex]
[tex]\[ -x^2 + 3x + 70 = 0 \][/tex]
- Rewriting in a standard quadratic form:
[tex]\[ x^2 - 3x - 70 = 0 \][/tex]
4. Solve the quadratic equation:
- To solve the quadratic equation [tex]\( x^2 - 3x - 70 = 0 \)[/tex], we can factorize it:
[tex]\[ (x - 10)(x + 7) = 0 \][/tex]
- Set each factor to zero:
[tex]\[ x - 10 = 0 \quad \Rightarrow \quad x = 10 \][/tex]
[tex]\[ x + 7 = 0 \quad \Rightarrow \quad x = -7 \][/tex]
5. Find the corresponding values of [tex]\( y \)[/tex]:
- If [tex]\( x = 10 \)[/tex], substitute back into [tex]\( y = 3 - x \)[/tex]:
[tex]\[ y = 3 - 10 = -7 \][/tex]
- If [tex]\( x = -7 \)[/tex], substitute back into [tex]\( y = 3 - x \)[/tex]:
[tex]\[ y = 3 + 7 = 10 \][/tex]
6. Verification:
- For [tex]\( x = 10 \)[/tex] and [tex]\( y = -7 \)[/tex]:
[tex]\[ x + y = 10 - 7 = 3 \][/tex]
[tex]\[ x \cdot y = 10 \cdot (-7) = -70 \][/tex]
- For [tex]\( x = -7 \)[/tex] and [tex]\( y = 10 \)[/tex]:
[tex]\[ x + y = -7 + 10 = 3 \][/tex]
[tex]\[ x \cdot y = -7 \cdot 10 = -70 \][/tex]
Thus, the two integers that add up to 3 and multiply to -70 are [tex]\( -7 \)[/tex] and [tex]\( 10 \)[/tex].
1. Let the two integers be [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
- We need to find [tex]\( x \)[/tex] and [tex]\( y \)[/tex] such that their sum is 3 and their product is -70.
- Therefore, we have two equations:
[tex]\[ x + y = 3 \quad \text{(1)} \][/tex]
[tex]\[ x \cdot y = -70 \quad \text{(2)} \][/tex]
2. Express [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex] from equation (1):
- From equation (1), we can express [tex]\( y \)[/tex] as:
[tex]\[ y = 3 - x \][/tex]
3. Substitute [tex]\( y \)[/tex] into equation (2):
- Substitute [tex]\( y \)[/tex] from the above expression into equation (2):
[tex]\[ x \cdot (3 - x) = -70 \][/tex]
- Simplify this equation to get a quadratic equation:
[tex]\[ 3x - x^2 = -70 \][/tex]
[tex]\[ -x^2 + 3x + 70 = 0 \][/tex]
- Rewriting in a standard quadratic form:
[tex]\[ x^2 - 3x - 70 = 0 \][/tex]
4. Solve the quadratic equation:
- To solve the quadratic equation [tex]\( x^2 - 3x - 70 = 0 \)[/tex], we can factorize it:
[tex]\[ (x - 10)(x + 7) = 0 \][/tex]
- Set each factor to zero:
[tex]\[ x - 10 = 0 \quad \Rightarrow \quad x = 10 \][/tex]
[tex]\[ x + 7 = 0 \quad \Rightarrow \quad x = -7 \][/tex]
5. Find the corresponding values of [tex]\( y \)[/tex]:
- If [tex]\( x = 10 \)[/tex], substitute back into [tex]\( y = 3 - x \)[/tex]:
[tex]\[ y = 3 - 10 = -7 \][/tex]
- If [tex]\( x = -7 \)[/tex], substitute back into [tex]\( y = 3 - x \)[/tex]:
[tex]\[ y = 3 + 7 = 10 \][/tex]
6. Verification:
- For [tex]\( x = 10 \)[/tex] and [tex]\( y = -7 \)[/tex]:
[tex]\[ x + y = 10 - 7 = 3 \][/tex]
[tex]\[ x \cdot y = 10 \cdot (-7) = -70 \][/tex]
- For [tex]\( x = -7 \)[/tex] and [tex]\( y = 10 \)[/tex]:
[tex]\[ x + y = -7 + 10 = 3 \][/tex]
[tex]\[ x \cdot y = -7 \cdot 10 = -70 \][/tex]
Thus, the two integers that add up to 3 and multiply to -70 are [tex]\( -7 \)[/tex] and [tex]\( 10 \)[/tex].
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