Question

How do you use the quadratic formula is used to find the roots of the equation `x^2-6x-19=0`?

Answer 1

`x_1=3-2sqrt 7`

`x_2=3+2sqrt 7`

Explanation:

`x^2-6x-19=0`

`"let us find out discriminant of the function "x^2-6x-19`

`Delta=sqrt(b^2-4ac)`

`a=1" , "b=-6" , "c=-19`

`Delta=sqrt((-6)^2-4*1*(-19)`

`Delta=sqrt(36+76)`

`Delta=sqrt(112)`

`Delta=+-4sqrt7`

`x_1=(-b-Delta)/(2a)=(6-4sqrt 7)/(2*1)=(6-4sqrt 7)/2=3-2sqrt 7`

`x_2=(-b+Delta)/(2a)=(6+4sqrt 7)/(2*1)=(6+4sqrt 7)/2=3+2sqrt 7`

Answer 2

Solution : `x=3+2sqrt7 , x= 3- 2sqrt7`

Explanation:

Comparing with general equation `ax^2+bx+c=0`

`x^2-6x-19=0 ; a=1 , b= -6 , c=-19`. Discriminant `D=b^2-4*a*c = 36+76=112` is positive so roots are real.

Quadratic formula for finding roots is`x= -b/(2a) +- sqrt(b^2-4ac)/(2a) or x = -(-6)/2+-sqrt112/2 or x = 3 +- 4*sqrt7/2 or x = 3+- 2*sqrt7`

Solution : `x=3+2sqrt7 , x= 3-2sqrt7` [Ans]