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

Question

Avery Alexander · Accepted Answer

#x=4+sqrt3#
Or
#x=4-sqrt3# PRIMER METHOD: By completing the square and using polynomial properties, one can find the zeros of a given polynomial. #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 provided algebraic expression's zeros are #-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# Secondly, The quadratic formula is used to find the zeros of the given polynomial. #-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#

Eli Assouline · Answer

undefined To find the zeros of the function $ f(x) = -2x^2 + 16x - 26 $ using the quadratic formula, first identify the coefficients $ a = -2 $, $ b = 16 $, and $ c = -26 $. Then apply the quadratic formula:

$$ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} $$

Substitute the values of $ a $, $ b $, and $ c $ into the formula:

$$ x = \frac{{-16 \pm \sqrt{{16^2 - 4(-2)(-26)}}}}{{2(-2)}} $$

$$ x = \frac{{-16 \pm \sqrt{{256 - 208}}}}{{-4}} $$

$$ x = \frac{{-16 \pm \sqrt{{48}}}}{{-4}} $$

$$ x = \frac{{-16 \pm 4\sqrt{{3}}}}{{-4}} $$

$$ x = \frac{{-16}}{{-4}} \pm \frac{{4\sqrt{{3}}}}{{-4}} $$

$$ x = 4 \mp \sqrt{{3}} $$

So, the zeros of the function are $ x = 4 + \sqrt{{3}} $ and $ x = 4 - \sqrt{{3}} $.