i have a test tomorrow and i cant remember how to do this!!!

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the square root of a number increased by twice the number is 8. find the number

Apr 29th, 2015

do you have any options?

if so i can answer accordingly

Apr 29th, 2015

√x + 2x = 8

√x= 8 - 2x

 x = (8 -2x)^2

 x = 64 - 32x + 4x ^2

4x^2  - 33x + 64  = 0                     (a=4 , b=  -33,c =64)

by solving

x=( -b +/- √(b^2 - 4ac)) / 2a

numbers are,

   

x1=33/8√ 65  / 8.....................................

x2=33/8−  √ 65 / 8...........................................







Apr 29th, 2015

i think the question is square of a number instead of square root of a number

if the question is

 the square  ( instead of  square root) of a number increased by twice the number is 8.

then ,

 x^2 + 2x = 8

 x^2 + 2x - 8 =0

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

(x + 4)= 0   or   (x -2) = 0

x= -4    or x = 2


numbers are  2, -4.......


Apr 29th, 2015

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Apr 29th, 2015
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