What is the roots for the following equation

Question

Can someone please show me to solve the roots for the following equation

$9x^2-8x-1 < 0 $

I am getting the root as below

$(x-9)(x+1)$ then getting $x = 9 ,~x = -1$ , which is wrong.

I have solved is as follows

Multiplication should give the value $-9$

Addition should give the value $-8$

So $(x-9)(x+1)$ is satisfying the above. Please tell me what am I doing wrong.

Answer 1

To find the roots of the quadratic polynomial $ax^2+bx+c$ we calculate the discriminant $$\Delta=b^2-4ac$$ and then the two roots are ($\Delta$ may be negative) $$x_{1,2}=\frac{-b\pm\sqrt{\Delta}}{2a}$$ and finally we factorize $$ax^2+bx+c=a(x-x_1)(x-x_2)$$

Answer 2

For Vieta formulas to apply, you have to divide by the coefficient of the highest power. So divide by $9$ and than the product of roots should be $\frac19$ while the sum is $\frac89$...