Question 114678
Let's use the quadratic formula to solve for z:



Starting with the general quadratic


{{{az^2+bz+c=0}}}


the general solution using the quadratic equation is:


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




So lets solve {{{9*z^2+12*z+4=0}}} ( notice {{{a=9}}}, {{{b=12}}}, and {{{c=4}}})





{{{z = (-12 +- sqrt( (12)^2-4*9*4 ))/(2*9)}}} Plug in a=9, b=12, and c=4




{{{z = (-12 +- sqrt( 144-4*9*4 ))/(2*9)}}} Square 12 to get 144  




{{{z = (-12 +- sqrt( 144+-144 ))/(2*9)}}} Multiply {{{-4*4*9}}} to get {{{-144}}}




{{{z = (-12 +- sqrt( 0 ))/(2*9)}}} Combine like terms in the radicand (everything under the square root)




{{{z = (-12 +- 0)/(2*9)}}} Simplify the square root (note: If you need help with simplifying the square root, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)




{{{z = (-12 +- 0)/18}}} Multiply 2 and 9 to get 18


So now the expression breaks down into two parts


{{{z = (-12 + 0)/18}}} or {{{z = (-12 - 0)/18}}}


Lets look at the first part:


{{{x=(-12 + 0)/18}}}


{{{z=-12/18}}} Add the terms in the numerator

{{{z=-2/3}}} Divide


So one answer is

{{{z=-2/3}}}




Now lets look at the second part:


{{{x=(-12 - 0)/18}}}


{{{z=-12/18}}} Subtract the terms in the numerator

{{{z=-2/3}}} Divide


So another answer is

{{{z=-2/3}}}


So our only solution is:

{{{z=-2/3}}}


Notice when we graph {{{9*x^2+12*x+4}}} (just replace z with x), we get:


{{{ graph( 500, 500, -12, 8, -12, 8,9*x^2+12*x+4) }}}


and we can see that the roots is {{{x=-2/3}}}. This verifies our answer