Question 88707


I'm assuming you can solve these using any technique you want right?


So let's use the quadratic formula to solve these quadratics


#1

Starting with the general quadratic


{{{ax^2+bx+c=0}}}


the general solution using the quadratic equation is:


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


So lets solve {{{x^2+3*x=0}}} (note: since the polynomial does not have a constant term, the 3rd coefficient is zero. In other words, c=0. So that means the polynomial really looks like {{{x^2+3*x+0=0}}}  notice {{{a=1}}}, {{{b=3}}}, and {{{c=0}}})


{{{x = (-3 +- sqrt( (3)^2-4*1*0 ))/(2*1)}}} Plug in a=1, b=3, and c=0




{{{x = (-3 +- sqrt( 9-4*1*0 ))/(2*1)}}} Square 3 to get 9




{{{x = (-3 +- sqrt( 9+0 ))/(2*1)}}} Multiply {{{-4*0*1}}} to get {{{0}}}




{{{x = (-3 +- sqrt( 9 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (-3 +- 3)/(2*1)}}} Simplify the square root




{{{x = (-3 +- 3)/2}}} Multiply 2 and 1 to get 2


So now the expression breaks down into two parts


{{{x = (-3 + 3)/2}}} or {{{x = (-3 - 3)/2}}}


Lets look at the first part:


{{{x=0/2}}} Add the terms in the numerator

{{{x=0}}} Divide


So one answer is

{{{x=0}}}

Now lets look at the second part:


{{{x=-6/2}}} Subtract the terms in the numerator

{{{x=-3}}} Divide


So another answer is

{{{x=-3}}}


So our solutions are:

{{{x=0}}} or {{{x=-3}}}


Notice when we graph {{{x^2+3*x}}}, we get:


{{{ graph( 500, 500, -13, 10, -13, 10,1*x^2+3*x+0) }}}


and we can see that the roots are {{{x=0}}} and {{{x=-3}}}. This verifies our answer


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#2



Starting with the general quadratic


{{{ax^2+bx+c=0}}}


the general solution using the quadratic equation is:


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


So lets solve {{{x^2+7*x+12=0}}} ( notice {{{a=1}}}, {{{b=7}}}, and {{{c=12}}})


{{{x = (-7 +- sqrt( (7)^2-4*1*12 ))/(2*1)}}} Plug in a=1, b=7, and c=12




{{{x = (-7 +- sqrt( 49-4*1*12 ))/(2*1)}}} Square 7 to get 49




{{{x = (-7 +- sqrt( 49+-48 ))/(2*1)}}} Multiply {{{-4*12*1}}} to get {{{-48}}}




{{{x = (-7 +- sqrt( 1 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (-7 +- 1)/(2*1)}}} Simplify the square root




{{{x = (-7 +- 1)/2}}} Multiply 2 and 1 to get 2


So now the expression breaks down into two parts


{{{x = (-7 + 1)/2}}} or {{{x = (-7 - 1)/2}}}


Lets look at the first part:


{{{x=-6/2}}} Add the terms in the numerator

{{{x=-3}}} Divide


So one answer is

{{{x=-3}}}

Now lets look at the second part:


{{{x=-8/2}}} Subtract the terms in the numerator

{{{x=-4}}} Divide


So another answer is

{{{x=-4}}}


So our solutions are:

{{{x=-3}}} or {{{x=-4}}}


Notice when we graph {{{x^2+7*x+12}}}, we get:


{{{ graph( 500, 500, -14, 7, -14, 7,1*x^2+7*x+12) }}}


and we can see that the roots are {{{x=-3}}} and {{{x=-4}}}. This verifies our answer