The set of all values of k for which the ineq | vedclass

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(C) For the quadratic expression $ax^2 + bx + c > 0$ to be true for all $x \in R$,we must have $a > 0$ and the discriminant $D = b^2 - 4ac < 0$.Here,$

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$(-\infty, -\frac{13}{7}) \cup (1, \infty)$

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(C) For the quadratic expression $ax^2 + bx + c > 0$ to be true for all $x \in R$,we must have $a > 0$ and the discriminant $D = b^2 - 4ac Here,$a = 1 > 0$,which is always true.The discriminant $D = \{-(3k + 1)\}^2 - 4(1)(4k^2 + 3k - 3) Expanding this,we get $(9k^2 + 6k + 1) - (16k^2 + 12k - 12) $-7k^2 - 6k + 13 Multiplying by $-1$ reverses the inequality: $7k^2 + 6k - 13 > 0$.Factoring the quadratic: $(7k + 13)(k - 1) > 0$.The roots are $k = -\frac{13}{7}$ and $k = 1$.For the expression to be positive,$k$ must lie outside the interval $[-\frac{13}{7}, 1]$.Thus,the solution set is $k \in (-\infty, -\frac{13}{7}) \cup (1, \infty)$.

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

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