Question 1085807


1. Solve the following quadratic equation:  

{{{2x^2-32=0}}}

{{{2x^2=32}}}

{{{x^2=32/2}}}

{{{x^2=16}}}

{{{x= sqrt(16)}}}

solutions:
{{{x=4}}}
{{{x=-4}}}


2. Given the function  {{{f(x)=x^2+12x}}}, which c-value could be added to both sides of the equation when solving by completing the square?

{{{f(x)+c^2=(x^2+12x+c^2)}}}...since {{{a=1}}} and {{{2ac=12}}}=>{{{c=6}}}

{{{f(x)+6^2=(x^2+12x+6^2)}}}

{{{f(x)+36=(x+6)^2}}}



3. Use the quadratic formula to solve.  

{{{4x^2-3x-5=0}}}

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

{{{x = (-(-3) +- sqrt( (-3)^2-4*4*(-5) ))/(2*4) }}}

{{{x = (3 +- sqrt( 9+80 ))/8 }}}

{{{x = (3 +- sqrt( 89 ))/8 }}}

{{{x = (3 +- sqrt( 89 ))/8 }}}

exact solutions:

{{{x = 3/8 + sqrt( 89 )/8 }}}

{{{x = 3/8 - sqrt( 89 )/8 }}}


4. Solve : 

{{{(x-1)^2-3=13}}}

{{{(x-1)^2=13+3}}}

{{{(x-1)^2=16}}}

{{{(x-1)= sqrt(16)}}}

{{{x= sqrt(16)+1}}}

solutions:

{{{x= 4+1}}}->{{{x=5}}}
{{{x= -4+1}}}->{{{x=-3}}}