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

2 Answers
May 3, 2017

Answer:

#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#

May 3, 2017

Answer:

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]