Question

The set of all values of \(k\) for which the inequality \(x^2 - (3k+1)x + 4k^2 + 3k - 3>0\) is true for all real values of \(x\) is

• A. \((-\frac{13}{7}, 1)\)
• B. \((-1, \frac{13}{7})\)
• C. \((-\infty, -\frac{13}{7}) \cup (1, \infty)\)
• D. \((-\infty, -1) \cup (\frac{13}{7}, \infty)\)

Hint:

A quadratic function is always positive if and only if its leading coefficient is positive, and its discriminant is negative.

Answer 1

The Correct Option is C

A quadratic function is always positive if and only if its leading coefficient is positive, and its discriminant is negative.The leading coefficient of \(x^2- (3k + 1) x + 4k^2 + 3k - 3\) is 1, which is positive.The discriminant is \[(3k + 1)^2 - 4 (4k^2 + 3k - 3) = 9k^2 + 6k + 1 - 16k^2 - 12k + 12 = -7k^2 - 6k + 13.\]We want this to be negative: \[-7k^2 - 6k + 13<0.\]We multiply both sides by $-1$, and reverse the direction of the inequality:\[7k^2 + 6k - 13>0.\]This factors as \((7k + 13)(k - 1)>0\).The roots are \(k = -\frac{13}{7}\) and \(k = 1\).We make a sign table: 
The solution is \(k<-\frac{13}{7}\) or \(k>1\), which can be written in interval notation as \(\left( -\infty, -\frac{13}{7} \right) \cup (1,\infty)\).