Question

Find the number of integral values of x which satisfy the inequality x₂–10x+19<6. 
 

• A. 5
• B. 11
• C. 7
• D. 8

Answer 1

The Correct Option is C

To solve the inequality \(x^2 - 10x + 19 < 6\), we will first simplify and solve the quadratic inequality step by step.

1. Simplify the inequality:
   • Start by subtracting 6 from both sides:
   • \(x^2 - 10x + 19 - 6 < 0\)
   • This reduces to \(x^2 - 10x + 13 < 0\).
2. Find the roots of the equation \(x^2 - 10x + 13 = 0\) by using the quadratic formula:

The quadratic formula is given by \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -10\), and \(c = 13\).

1. Substitute these values into the formula:
   • \(x = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(1)(13)}}{2(1)}\)
   • \(x = \frac{10 \pm \sqrt{100 - 52}}{2}\)
   • \(x = \frac{10 \pm \sqrt{48}}{2}\)
   • \(x = \frac{10 \pm 4\sqrt{3}}{2}\)
   • \(x = 5 \pm 2\sqrt{3}\)
2. Calculate the approximate numerical values of the roots:
   • \(x_1 = 5 + 2\sqrt{3} \approx 8.464\)
   • \(x_2 = 5 - 2\sqrt{3} \approx 1.536\)
3. These values split the real number line into intervals. We need to check in which intervals the original inequality holds true:
   • Consider the intervals: \((-\infty, 1.536)\), \((1.536, 8.464)\), \((8.464, \infty)\).
   • Test a value from each interval to determine where the inequality \(x^2 - 10x + 13 < 0\) holds:
     • Choose \(x = 0\) in the first interval:
       • \(0^2 - 10(0) + 13 = 13 > 0\)
     • Choose \(x = 2\) in the second interval:
       • \(2^2 - 10(2) + 13 = -3 < 0\)
     • Choose \(x = 9\) in the third interval:
       • \(9^2 - 10(9) + 13 = 4 > 0\)
4. Thus, the inequality \(x^2 - 10x + 13 < 0\) is satisfied in the interval \((1.536, 8.464)\).
5. Determine the integral values of \(x\) in this interval:
   • The integer values of \(x\) that lie in the interval \((1.536, 8.464)\) are \(2, 3, 4, 5, 6, 7,\) and \(8\).

Therefore, there are 7 integral values of \(x\) that satisfy the inequality \(x^2 - 10x + 19 < 6\).