Question 96217
{{{4w^2=10w-4}}}



{{{4w^2-10w+4=0}}} Get all terms to one side




Let's use the quadratic formula to solve for w:



Starting with the general quadratic


{{{aw^2+bw+c=0}}}


the general solution using the quadratic equation is:


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


So lets solve {{{4*w^2-10*w+4=0}}} ( notice {{{a=4}}}, {{{b=-10}}}, and {{{c=4}}})


{{{w = (--10 +- sqrt( (-10)^2-4*4*4 ))/(2*4)}}} Plug in a=4, b=-10, and c=4




{{{w = (10 +- sqrt( (-10)^2-4*4*4 ))/(2*4)}}} Negate -10 to get 10




{{{w = (10 +- sqrt( 100-4*4*4 ))/(2*4)}}} Square -10 to get 100  (note: remember when you square -10, you must square the negative as well. This is because {{{(-10)^2=-10*-10=100}}}.)




{{{w = (10 +- sqrt( 100+-64 ))/(2*4)}}} Multiply {{{-4*4*4}}} to get {{{-64}}}




{{{w = (10 +- sqrt( 36 ))/(2*4)}}} Combine like terms in the radicand (everything under the square root)




{{{w = (10 +- 6)/(2*4)}}} 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>)




{{{w = (10 +- 6)/8}}} Multiply 2 and 4 to get 8


So now the expression breaks down into two parts


{{{w = (10 + 6)/8}}} or {{{w = (10 - 6)/8}}}


Lets look at the first part:


{{{x=(10 + 6)/8}}}


{{{w=16/8}}} Add the terms in the numerator

{{{w=2}}} Divide


So one answer is

{{{w=2}}}




Now lets look at the second part:


{{{x=(10 - 6)/8}}}


{{{w=4/8}}} Subtract the terms in the numerator

{{{w=1/2}}} Divide


So another answer is

{{{w=1/2}}}


So our solutions are:

{{{w=2}}} or {{{w=1/2}}}


Notice when we graph {{{4*x^2-10*x+4}}} (just replace w with x), we get:


{{{ graph( 500, 500, -9, 12, -9, 12,4*x^2+-10*x+4) }}}


and we can see that the roots are {{{w=2}}} and {{{w=1/2}}}. This verifies our answer