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Answer :
To solve the problem, we need to find which equation correctly represents a situation where two positive integers multiply to give a product of 176. We are specifically looking for the equation that models finding the greater of these two integers.
Let's examine each option:
1. [tex]\( x^2 + 5 = 176 \)[/tex]
- Solving for [tex]\( x \)[/tex]:
[tex]\[
x^2 + 5 = 176 \\
x^2 = 171 \\
x = \sqrt{171}
\][/tex]
- [tex]\( \sqrt{171} \)[/tex] is not an integer, so this option does not yield a positive integer.
2. [tex]\( x(x + 5) = 176 \)[/tex]
- Expanding gives [tex]\( x^2 + 5x = 176 \)[/tex], which is not directly usable for our needs, but solving for [tex]\( x \)[/tex] if possible:
[tex]\[
x^2 + 5x = 176
\][/tex]
- Trying different values, we see that substituting with integers can verify or require more work as a quadratic with exact values.
3. [tex]\( x(x - 5) = 176 \)[/tex]
- Expanding gives [tex]\( x^2 - 5x = 176 \)[/tex].
- Rearrange: [tex]\( x^2 - 5x - 176 = 0 \)[/tex].
- Solve this quadratic equation (can use the quadratic formula):
[tex]\[
a = 1, \quad b = -5, \quad c = -176 \\
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\
x = \frac{5 \pm \sqrt{25 + 704}}{2} \\
x = \frac{5 \pm \sqrt{729}}{2} \\
x = \frac{5 \pm 27}{2}
\][/tex]
- Solutions: [tex]\( x = \frac{32}{2} = 16 \)[/tex] or [tex]\( x = \frac{-22}{2} = -11 \)[/tex].
- The positive solution is [tex]\( x = 16 \)[/tex].
4. [tex]\( x^2 - 5 = 176 \)[/tex]
- Solving for [tex]\( x \)[/tex]:
[tex]\[
x^2 - 5 = 176 \\
x^2 = 181 \\
x = \sqrt{181}
\][/tex]
- [tex]\( \sqrt{181} \)[/tex] isn't an integer, so this doesn't work.
Conclusively, by solving option 3, we find that the greater positive integer that multiplies by another to give 176 is [tex]\( x = 16 \)[/tex].
Let's examine each option:
1. [tex]\( x^2 + 5 = 176 \)[/tex]
- Solving for [tex]\( x \)[/tex]:
[tex]\[
x^2 + 5 = 176 \\
x^2 = 171 \\
x = \sqrt{171}
\][/tex]
- [tex]\( \sqrt{171} \)[/tex] is not an integer, so this option does not yield a positive integer.
2. [tex]\( x(x + 5) = 176 \)[/tex]
- Expanding gives [tex]\( x^2 + 5x = 176 \)[/tex], which is not directly usable for our needs, but solving for [tex]\( x \)[/tex] if possible:
[tex]\[
x^2 + 5x = 176
\][/tex]
- Trying different values, we see that substituting with integers can verify or require more work as a quadratic with exact values.
3. [tex]\( x(x - 5) = 176 \)[/tex]
- Expanding gives [tex]\( x^2 - 5x = 176 \)[/tex].
- Rearrange: [tex]\( x^2 - 5x - 176 = 0 \)[/tex].
- Solve this quadratic equation (can use the quadratic formula):
[tex]\[
a = 1, \quad b = -5, \quad c = -176 \\
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\
x = \frac{5 \pm \sqrt{25 + 704}}{2} \\
x = \frac{5 \pm \sqrt{729}}{2} \\
x = \frac{5 \pm 27}{2}
\][/tex]
- Solutions: [tex]\( x = \frac{32}{2} = 16 \)[/tex] or [tex]\( x = \frac{-22}{2} = -11 \)[/tex].
- The positive solution is [tex]\( x = 16 \)[/tex].
4. [tex]\( x^2 - 5 = 176 \)[/tex]
- Solving for [tex]\( x \)[/tex]:
[tex]\[
x^2 - 5 = 176 \\
x^2 = 181 \\
x = \sqrt{181}
\][/tex]
- [tex]\( \sqrt{181} \)[/tex] isn't an integer, so this doesn't work.
Conclusively, by solving option 3, we find that the greater positive integer that multiplies by another to give 176 is [tex]\( x = 16 \)[/tex].
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