Determine the values for which \(x^2 - 7x + 10 \leq 0\)

A.

2 \(\leq\) x \(\geq\) 5

B.

-2 \(\leq\) x \(\leq\) 3

C.

-2 \(\leq\) x \(\geq\) 3

D.

2 \(\leq\) x \(\leq\) 5

Correct answer is 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.