Question

If 'x' is a number such that \(x^2 - 5x + 4<0\) and \(x^2 - 3x + 2<0\), which of the following can be the value of 'x'?

• A. 3.5
• B. 3.0
• C. 2.4
• D. 1.6
• E. 0.8

Hint:

To solve a quadratic inequality like \(ax^2+bx+c<0\) (with \(a>0\)), find the roots \(r_1\) and \(r_2\). The solution will be the interval between the roots: \(r_1<x<r_2\). If the inequality is \(>0\), the solution will be the intervals outside the roots: \(x<r_1\) or \(x>r_2\).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
We need to find a value of 'x' that simultaneously satisfies two quadratic inequalities. The process is to solve each inequality separately and then find the intersection of their solution sets.
Step 2: Detailed Explanation:
Solve the first inequality: \(x^2 - 5x + 4<0\)
First, factor the quadratic expression: \[ (x - 1)(x - 4)<0 \] The roots of the equation \((x-1)(x-4)=0\) are \(x=1\) and \(x=4\). Since the parabola opens upward, the expression is negative (less than 0) between the roots. The solution for the first inequality is \(1<x<4\).
Solve the second inequality: \(x^2 - 3x + 2<0\)
First, factor the quadratic expression: \[ (x - 1)(x - 2)<0 \] The roots of the equation \((x-1)(x-2)=0\) are \(x=1\) and \(x=2\). Since the parabola opens upward, the expression is negative between the roots. The solution for the second inequality is \(1<x<2\).
Find the intersection of the solutions.
We need a value of 'x' that satisfies both \(1<x<4\) AND \(1<x<2\). The intersection of these two intervals is \(1<x<2\).
Check the options.
We must find which of the given options lies in the interval (1, 2).

(A) 3.5 is not in (1, 2).
(B) 3.0 is not in (1, 2).
(C) 2.4 is not in (1, 2).
(D) 1.6 is in (1, 2).
(E) 0.8 is not in (1, 2).
Step 3: Final Answer:
The only value that satisfies both inequalities is 1.6. This corresponds to option (D).