How do you solve 16x^2<15x+1?

Dec 30, 2016

The answer is x in ] -1/16 , 1 [

Explanation:

Let's rewrite the equation

$16 {x}^{2} - 15 x - 1 < 0$

Let $f \left(x\right) = 16 {x}^{2} - 15 x - 1$

The domain of $f \left(x\right)$ is ${D}_{f} \left(x\right) = \mathbb{R}$

We can factorise the RHS

$16 {x}^{2} - 15 x - 1 = \left(16 x + 1\right) \left(x - 1\right)$

Now we can establish the sign chart

$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a a}$$- \frac{1}{16}$$\textcolor{w h i t e}{a a a a a}$$1$$\textcolor{w h i t e}{a a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$16 x + 1$$\textcolor{w h i t e}{a a a a a a a a}$$-$$\textcolor{w h i t e}{a a a a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$x - 1$$\textcolor{w h i t e}{a a a a a a a a a a}$$-$$\textcolor{w h i t e}{a a a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$f \left(x\right)$$\textcolor{w h i t e}{a a a a a a a a a a a}$$+$$\textcolor{w h i t e}{a a a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$

Therefore,

$f \left(x\right) < 0$, when x in ] -1/16 , 1 [