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

do you have any options?

if so i can answer accordingly

√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...........................................

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.......

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