If kx2−4x+3k+1&gt;0 for at least one x&gt;0 and if k∈S, find S

If&nbsp;k≥0, then the said condition will always be true, as we get an upward facing parabola. (or a line, for the&nbsp;k=0&nbsp;case)If&nbsp;k&lt;0, then we need the discriminant to be&nbsp;&gt;0, for otherwise the quadratic always assumes negative values, i.e.16−4k(3k+1)&gt;0⇒k∈(−43,1)Now, since the quadratic must be positive for at least one positive x, at least one of the roots must be positive (can you see why?), so4+16−4k(3k+1)2k&gt;0But this is never true for&nbsp;k&lt;0. Although4−16−4k(3k+1)2k&gt;0is true for&nbsp;k∈(−13,0)Taking the union of the two cases, we must have thatk∈(-13,∞)

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High School
Algebra
Algebra I
Quadratic function and equation

Benjamin Hampton

Answered question

2022-04-21

k x^{2} - 4 x + 3 k + 1 > 0

for at least one

x > 0

and if

k \in S

, find S

Answer & Explanation

Cynthia Herrera

Beginner2022-04-22Added 16 answers

If $k \geq 0$ , then the said condition will always be true, as we get an upward facing parabola. (or a line, for the $k = 0$ case)
If $k < 0$ , then we need the discriminant to be $> 0$ , for otherwise the quadratic always assumes negative values, i.e.
$16 - 4 k (3 k + 1) > 0$

$\Rightarrow k \in (- \frac{4}{3}, 1)$
Now, since the quadratic must be positive for at least one positive x, at least one of the roots must be positive (can you see why?), so
$\frac{4 + \sqrt{16 - 4 k (3 k + 1)}}{2 k} > 0$
But this is never true for $k < 0$ . Although
$\frac{4 - \sqrt{16 - 4 k (3 k + 1)}}{2 k} > 0$
is true for $k \in (- \frac{1}{3}, 0)$
Taking the union of the two cases, we must have that
$k \in (- \frac{1}{3}, \infty)$

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