Question

For what values of k will the function f(x) = 9x² + 4x + k have 1 real roots.

Answer 1

Answer:

$\{\frac{4}{9} \}$

Step-by-step explanation:

First off, we are given a parabola by definition:

$f(x) = 9x^2 + 4x + k \\ a = 9, b = 4, c = k;$

Since $a > 0$, our parabola has an upward opening.

Should we think about it graphically, it will be no wonder that we should pay our attention to the points of the parabola intersecting the abscissa axis. In other words, we need our vertex to intersect the x axis only once per se.
The vertex:

$V_{x} = \frac{-b}{2a} = \frac{-4}{2 * 9} = - \frac{2}{9} \\ V_{y} = 9V_{x}^2 + 4V_{x} + k = 9(- \frac{2}{9})^2 + 4(- \frac{2}{9}) + k = 9 * \frac{4}{9^2} - \frac{8}{9} + k = \frac{4}{9} - \frac{8}{9} + k = - \frac{4}{9} + k$

As a result, our $V_{y}$ is parametric. $k = \frac{4}{9}$ suits us because $V_{y} = - \frac{4}{9} + \frac{4}{9} = 0$. If $k > \frac{4}{9}$, we do not have any roots at all. We have two roots if $k < \frac{4}{9}$.

Answer 2

Hello,

Answer:

for k = 4/9

Step-by-step explanation:

a = 9 ; b = 4 ; c = k

we search Δ = 0 (because 1 real root)

Δ = b² - 4ac = 4² - 4 × 9 × k = 16 - 36k

Δ = 0 ⇔  16 - 36k = 0 ⇔ 36k = 16 ⇔ k = 16/36 = (4 × 4)/(4 × 9) = 4/9

⇒ k = 4/9