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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.

RandomUser
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Gaurav_soni
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2 Answers2

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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 $$ax^2+bx+c=a(x-x_1)(x-x_2)$$

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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 $\frac89$...

draks ...
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  • but the method i am using works for almost all the equations with some exceptions. Even on sites they tell to do using the same method. When should we be dividing the by the number of the highest power ? – Gaurav_soni Apr 29 '14 at 05:58
  • @gaurav what are the exceptions? You should always divide! Then you conquer the equations... – draks ... Jun 03 '14 at 08:22