Given #-2x^2+16x-26# how do you use the quadratic formula to find the zeros of f(x)?

1 Answer
Nov 1, 2016

Answer:

#x=4+sqrt3#
Or
#x=4-sqrt3#

Explanation:

FIRST METHOD:

The zeros of the given polynomial is determined by completing the square and applying polynomial properties

#color(brown)((a-b)^2=a^2-2ab+b^2)#
#color(violet)(a^2-b^2=(a-b)(a+b))#

#-2x^2+16x-26#

#=color(blue)(-2)(color(blue)1x^2color(blue)(-8)xcolor(blue)(+13))#

#=-2(x^2-8x+13color(red)(+3-3))#

#=-2(x^2-8x+16-3)#

#=-2(color(brown)(x^2-2(4)x+4^2)-3)#

#=-2(color(brown)((x-4)^2)-3)^#

#=-2color(violet)(((x-4)^2-(sqrt3)^2))#

#=-2color(violet)(((x-4)-sqrt3)((x-4)+sqrt3))#

#=-2((x-(4+sqrt3)((x-(4-sqrt3))#

The zeros of the given algebraic expression is

#-2x^2+16x-26=0#

#-2((x-(4+sqrt3)((x-(4-sqrt3))=0#

#x-(4+sqrt3)=0rArrx=4+sqrt3#
Or
#x-(4-sqrt3)=0rArrx=4-sqrt3#

SECOND METHOD:

The zeros of the given polynomial is determined by using the quadratic formula

#-2x^2+16x-26#

#color(blue)(delta=b^2-4ac)=16^2-4(-2)(-26)#
#delta=256-208=48#

Roots are:
#color(red)(x_1=(-b-sqrtdelta)/(2a))=(-16-sqrt48)/(2(-2))=(-16-sqrt(2^4xx3))/(-4)#
#color(red)(x_1)=(-16-4sqrt3)/(-4)=(-4(4+sqrt3))/(-4)=(cancel(-4)(4+sqrt3))/cancel(-4)#

#color(red)(x_1)=4+sqrt3#

#color(red)(x_2=(-b+sqrtdelta)/(2a))=(-16+sqrt48)/(2(-2))=(-16+sqrt(2^4xx3))/(-4)#
#color(red)(x_2)=(-16+4sqrt3)/(-4)=(-4(4-sqrt3))/(-4)=(cancel(-4)(4-sqrt3))/cancel(-4)#

#color(red)(x_2)=4-sqrt3#