contestada

when the square of a number is added to the number trebled, the result is 108. what is the number? (solve it in quadratic context, plsss, finding it really difficult)

Respuesta :

caylus
Hello,

Let's assume x the number

x^2+3x=108
==>x^2+3x-108=0
Δ=3²+4*108=441=21²

x= (-3+21)/2 or x=(-3-21)/2
==>x=9 or x=-12

apologiabiology
number is 'n'

square of a number means n^2
added to means plus or +
number trippled is 3n
result means equals or =
108 is 108


n^2+3n=108
minus 108 from both sides
n^2+3n-108=0
if want quadratic formula
in form
ax^2+bx+c=0
x=[tex] \frac{-b+/- \sqrt{b^2-4ac} }{2a} [/tex]
a=1
b=3
c=-108


x=[tex] \frac{-3+/- \sqrt{3^2-4(1)(-108)} }{2(1)} [/tex]
x=[tex] \frac{-3+/- \sqrt{9+432} }{2} [/tex]
x=[tex] \frac{-3+/- \sqrt{441} }{2} [/tex]
x=[tex] \frac{-3+/- 21 }{2} [/tex]
x=[tex] \frac{-3+21 }{2} [/tex] or x=[tex] \frac{-3- 21}{2} [/tex]
x=[tex] \frac{18 }{2} [/tex] or x=[tex] \frac{-24}{2} [/tex]
x=9 or -12

the number is either 9 or -12

Otras preguntas