Question 1083315

let's a number be {{{x}}}
twice its square is {{{2x^2}}}

i the sum of number and twice its square is {{{35}}}, we have

{{{2x^2+x=35}}}

 {{{2x^2+x-35=0}}}......use quadratic formula

{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 

{{{x = (-1 +- sqrt( 1^2-4*2*(-35) ))/(2*2) }}} 

{{{x = (-1 +- sqrt( 1+280 ))/4 }}} 

{{{x = (-1 +- sqrt( 281 ))/4 }}} 

{{{x = (-1 +- 16.76)/4 }}} 

solutions:

{{{x = (-1 + 16.76)/4 }}} 

{{{x =  15.76/4 }}} 

{{{x =  3.94}}} 

or

{{{x = (-1 - 16.76)/4 }}} 

{{{x = -17.76/4 }}} 

{{{x =  -4.44}}} 


{{{ graph( 600, 600, -10, 10, -40, 10, 2x^2+x-35) }}}