Question #84aa9

Question

986 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-01T02:30:41

#x = (-13+-3sqrt(17))/2#

#x^2 = -13x - 4#
Subtract #-13x - 4# from both sides:
#x^2 + 13x + 4 = 0#

Since it doesn't neatly factor, one of the simplest ways to calculate it is the quadratic formula:

#x = (-b+-sqrt(b^2-4ac))/(2a)# where #a# is the coefficient of #x^2#, #b# is the coefficient of #x#, #c# is the constant.

Substitute the values for #a (1)#, #b (13)#, and# c (4)#...

#x = (-(13)+-sqrt((13)^2-4(1)(4)))/(2(1))#

#x = (-13+-sqrt(153))/2#

#sqrt(153)# can also be represented as #3*sqrt(17)# (because

#sqrt(153) = sqrt(9 * 17) = sqrt(3^2 * 17)#), so it could also be shown as

#x = (-13+-3sqrt(17))/2#

Answer 2 · 2018-02-01T02:32:32

#x = color(blue)(-0.3153), color(red)(-12.6847#

Given : #x^2 + 13x +4 = 0#

Roots of a quadratic equation given by the formula,

#x = (-b +- sqrt(b^2 - 4ac)) / (2a)# where

Coefficient of term #x^2 = a#, coefficient of term #x = b# and coefficient of constant term = c

#:. x = (-13 +- sqrt(13^2 - (4 * 1 * 4))) / (2 * 1)#

#x = (-13 +- sqrt153)/2 = color(blue)(-0.3153), color(red)(-12.6847#