Question 1085653

I. Given the following polynomial: 
{{{2x^2 + 7x - 15 = 0 }}}

Check all that apply.
{{{b^2-4ac=7^2-4*2(-15)=49+120=169}}}
 The value of the discriminant is {{{169}}}.

 There are {{{2 }}}real roots.
{{{discriminant>0}}}, so there are 2 real roots

 The parabola is directed upward.
since {{{a=2}}} which is positive number, the  parabola{{{ is}}} directed upward

 The axis of symmetry is located at: {{{x = -7/4}}}

{{{2 (x + 7/4)^2 - 169/8 = 0}}}-> {{{h=-7/4}}}->the axis of symmetry goes through vertex ({{{-7/4}}},{{{- 169/8}}}) and it is a vertical line {{{x = -7/4}}}


 The graph intersects the y axis at ({{{0}}},{{{ -15}}}).
{{{y=2x^2 + 7x - 15 }}}
{{{y=2*0^2 + 7x* - 15}}}
{{{y=-15}}}

 The graph intersects the x-axis at ({{{-5}}}, {{{0}}})  and ({{{1.5}}},{{{ 0}}}) 
{{{(x + 5) (2 x - 3) = 0}}}
{{{(x + 5)  = 0}}} -> {{{x=-5}}} -> point ({{{-5}}}, {{{0}}}) 
{{{(2 x - 3) = 0 }}}-> {{{x=3/2=1.5 }}}-> ({{{1.5}}},{{{ 0}}}) 


{{{ graph( 600, 600, -10, 10, -20, 10, 2x^2 + 7x - 15) }}}


II.	
Use the quadratic formula to solve the following: 

{{{3x^2 - x + 2 = 0 }}}

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

{{{x = (-(-1) +- sqrt( (-1)^2-4*3*2 ))/(2*3) }}} 

{{{x = (1 +- sqrt(1-24 ))/6 }}} 

{{{x = (1 +- sqrt(-23 ))/6 }}} 

{{{x = (1 +- i*sqrt(23 ))/6 }}} 

solutions:

{{{x = (1/6)(1 + i*sqrt(23 )) }}} 

{{{x = (1/6)(1 - i*sqrt(23 )) }}}