Question 111423
#1



*[Tex \LARGE 9(x-8)^2=36] Start with the given equation





{{{(x-8)^2=4}}} Divide both sides by 9




*[Tex \LARGE x-8=\pm sqrt{4}] Take the square root of both sides





*[Tex \LARGE x-8=\pm 2] 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>)




*[Tex \LARGE x=8\pm 2] Add 8 to both sides to isolate x.



 Break down the expression into two parts:




 <pre>

 *[Tex \LARGE x=8+2]  <font size="6">or</font>  *[Tex \LARGE x=8-2]

 </pre>


 Now combine like terms for each expression:

 <pre>

 *[Tex \LARGE x=10]  <font size="6">or</font>   *[Tex \LARGE x=6]   </pre>



-----------------------------------

Answer:

So our solution is

 <pre>

*[Tex \LARGE x=10]  <font size="6">or</font>   *[Tex \LARGE x=6]

 </pre>


 Notice when we graph the equations {{{y=9(x-8)^2}}} and {{{y=36}}}  we get:


{{{drawing(500, 500, 1, 15, -10, 41,
graph( 500, 500, 1, 15, -10, 41, 9(x-8)^2,36)
)}}} graph of  {{{y=9(x-8)^2}}} (red) and {{{y=36}}} (green)




Here we can see that the two equations intersect at x values of {{{x=10}}} and {{{x=6}}}, so this verifies our answer.

 
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#2


*[invoke quadratic_formula 27, 0, -49, "x"]


<hr>



#3

 {{{ -16x = -x^2 }}} Start with the given equation



 {{{ x^2-16x =0  }}}Add {{{x^2}}} to both sides



{{{x(x-16)=0}}} Factor the left side (note: if you need help with factoring, check out this <a href=http://www.algebra.com/algebra/homework/playground/change-this-name4450.solver>solver</a>)




Now set each factor equal to zero:


{{{x=0}}} or {{{x-16=0}}}


{{{x=0}}} or {{{x=16}}}  Now solve for x in each case



So our solutions are {{{x=0}}} or {{{x=16}}}



Notice if we graph {{{y=x^2-16x}}} we get


{{{ graph(500,500,-20,20,-10,10, x^2-16x) }}}


and we can see that the graph has roots at {{{x=0}}} or {{{x=16}}}, so this verifies our answer.


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#4


{{{x^3-4x^2+4x=0}}} Start with the given expression



{{{x(x^2-4x+4)=0}}} Factor out the GCF {{{x}}}



{{{x(x-2)(x-2)=0}}} Factor the inner expression {{{x^2-4x+4}}}


Now set each factor equal to zero:


{{{x=0}}}, {{{x-2=0}}} or {{{x-2=0}}}


{{{x=0}}}, {{{x=2}}}, or {{{x=2}}}   Now solve for x in each case


Notice we get the solution {{{x=2}}} twice



So our solutions are {{{x=0}}} or {{{x=2}}}



Notice if we graph {{{y=x^3-4x^2+4x}}} we get


{{{ graph(500,500,-10,10,-10,10, x^3-4x^2+4x) }}}


and we can see that the graph has roots at {{{x=0}}} and {{{x=2}}} so this verifies our answer.



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#5


*[invoke quadratic_formula 1, 0, 69, "x"]