Question 1085809

1. Use the discriminant to determine the number of solutions for 

{{{x^2-3x+6=0}}}

{{{b^2-4ac}}}
{{{(-3)^2-4*1*6}}}
{{{9-24}}}
{{{-15}}} =>  if {{{discriminant<0}}} the quadratic equation will have {{{no}}} {{{real}}} solutions


2. Solve {{{x^2+6x=0}}}; factor

{{{x(x+6)=0}}}

solutions:

{{{x=0}}} and
{{{(x+6)=0}}}->{{{x=-6}}}


3. Solve the following quadratic equation using the specified method. (5 points per solution method)

{{{x^2+10x=3}}}
A) Use the method of completing the square and square roots.

{{{x^2+10x=3}}}

{{{(x^2+10x+b^2)-b^2=3}}}....since {{{a=1}}} and {{{2ab=10}}}->{{{b=5}}}

{{{(x^2+10x+5^2)-5^2=3}}}

{{{(x+5)^2-25=3}}}

{{{(x+5)^2=3+25}}}

{{{(x+5)^2=28}}}

{{{sqrt((x+5)^2)=sqrt(28)}}}

{{{x+5=sqrt(4*7)}}}

{{{x+5=2sqrt(7)}}}

{{{x=2sqrt(7)-5}}}

solutions:

{{{x=2sqrt(7)-5}}}

{{{x=-2sqrt(7)-5}}} 

B) Use the Quadratic Formula.

{{{x^2+10x=3}}}

{{{x^2+10x-3=0}}}

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

{{{x = (-10 +- sqrt(10^2-4*1*(-3)))/(2*1) }}} 

{{{x = (-10 +- sqrt(100+12))/2 }}} 

{{{x = (-10 +- sqrt(112))/2 }}} 

{{{x = (-cross(10)5 +- cross(4)2sqrt(7))/cross(2) }}} 

{{{x = (-5 +- 2sqrt(7)) }}} 

solutions:

{{{x = -5 + 2sqrt(7) }}} 

{{{x = -5 - 2sqrt(7) }}}