Question 1209290
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Consider the equation {{{x*(x+6) = 16 - x^2 + 14x}}}


Let's get everything to one side.
{{{x*(x+6) = 16 - x^2 + 14x}}}
{{{x^2+6x = 16 - x^2 + 14x}}}
{{{x^2+6x - 16 + x^2 - 14x = 0}}}
{{{2x^2 - 8x - 16 = 0}}}
{{{2(x^2 - 4x - 8) = 0}}}
{{{x^2 - 4x - 8 = 0}}}


That last equation is of the format  {{{ax^2+bx+c = 0}}}
where,
a = 1, b = -4, c = -8
Plug those into the quadratic formula.
{{{x = (-b +- sqrt(b^2 - 4ac))/(2a)}}}


{{{x = (-(-4) +- sqrt((-4)^2 - 4(1)(-8)))/(2(1))}}}


{{{x = (4 +- sqrt(16+32))/(2(1))}}}


{{{x = (4 +- sqrt(48))/(2)}}}


{{{x = (4 +- sqrt(16*3))/(2)}}}


{{{x = (4 +- sqrt(16)*sqrt(3))/(2)}}}


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


{{{x = (2(2 +- 2*sqrt(3)))/(2)}}}


{{{x = 2 +- 2*sqrt(3)}}}


{{{x = 2 + 2*sqrt(3)}}} or {{{x = 2 - 2*sqrt(3)}}}


{{{x = 5.464}}} or {{{x = -1.464}}} both of which are approximate.


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Draw a number line with regions A, B, C
region A = stuff to the left of -1.464
region B = stuff between -1.464 and 5.464
region C = stuff to the right of 5.464


Possible test values for regions A,B,C could be: x = -2, x = 0, x = 6 in that order
You plug each test value back into the original inequality {{{x*(x+6) > 16 - x^2 + 14x}}} to see if we get a true statement or not.


If we tried x = -2, then,
{{{x*(x+6) > 16 - x^2 + 14x}}}
{{{-2*(-2+6) > 16 - (-2)^2 + 14*(-2)}}}
{{{-8 > -16}}}
which is true.
Any x value in region A is in the solution set for {{{x*(x+6) > 16 - x^2 + 14x}}}


Region A is {{{x < 2 - 2*sqrt(3)}}} aka {{{-infinity < x < 2 - 2*sqrt(3)}}}
Yielding the interval notation {{{(matrix(1,3,-infinity,",",2-2sqrt(3)))}}}
Use curved parenthesis to indicate we do <u>not</u> include the endpoints in the solution set.


If we tried x = 0 then
{{{x*(x+6) > 16 - x^2 + 14x}}}
{{{0*(0+6) > 16 - 0^2 + 14*0}}}
{{{0 > 16}}}
which is false. 
Any x value in region B, ie the region between {{{x = 2 - 2*sqrt(3) = -1.464}}} and {{{x = 2 + 2*sqrt(3) = 5.464}}} will make the original inequality false.



The last region to test is region C.
Let's plug in x = 6
{{{x*(x+6) > 16 - x^2 + 14x}}}
{{{6*(6+6) > 16 - 6^2 + 14*6}}}
{{{72 > 64}}}
which is true
Therefore {{{x > 2 + 2*sqrt(3)}}} aka {{{2 + 2*sqrt(3) < x < infinity}}} translates to the interval notation {{{(matrix(1,3,2 + 2sqrt(3),",", infinity))}}}



We found these two interval notation regions
{{{(matrix(1,3,-infinity,",",2-2sqrt(3)))}}} or {{{(matrix(1,3,2 + 2sqrt(3),",", infinity))}}}


Glue them together with a set union symbol to end up with this <font color=red>final answer</font>
{{{matrix(1,3,(matrix(1,3,-infinity,",",2-2sqrt(3))),"U",(matrix(1,3,2 + 2sqrt(3),",", infinity)))}}}
You can use a graphing calculator like <a href="https://www.desmos.com/calculator">Desmos</a> and <a href="https://www.geogebra.org/calculator">GeoGebra</a> to confirm.
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