Determine the values for which \(x^2 - 7x + 10 \leq 0\)
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Correct Answer: Option D
Explanation:
\(x^2 - 7x + 10 \leq 0\)
Solve for \(x^2 - 7x + 10 = 0\)
We have, (x - 5)(x - 2) \(\leq\) 0.
Conditions:
Case 1: (x - 5) \(\leq\) 0, (x - 2) \(\geq\) 0.
\(\implies\) x \(\leq\) 5; x \(\geq\) 2.
2 \(\leq\) x \(\leq\) 5.
Choosing x = 3,
3\(^2\) - 7(3) + 10 = 9 - 21 + 10
= -2 \(\leq\) 0.
\(\therefore\) 2 \(\leq\) x \(\leq\) 5.
\(x^2 - 7x + 10 \leq 0\)
Solve for \(x^2 - 7x + 10 = 0\)
We have, (x - 5)(x - 2) \(\leq\) 0.
Conditions:
Case 1: (x - 5) \(\leq\) 0, (x - 2) \(\geq\) 0.
\(\implies\) x \(\leq\) 5; x \(\geq\) 2.
2 \(\leq\) x \(\leq\) 5.
Choosing x = 3,
3\(^2\) - 7(3) + 10 = 9 - 21 + 10
= -2 \(\leq\) 0.
\(\therefore\) 2 \(\leq\) x \(\leq\) 5.