Question 283152
{{{(2x^2+6x-3)/(x^2+x-13)>=1}}}
<pre><font size = 4 color = "indigo"><b>
Subtract 1 from both sides to get 0 on the right

{{{(2x^2+6x-3)/(x^2+x-13)-1>=0}}}

Write {{{1}}} as {{{1/1}}}

{{{(2x^2+6x-3)/(x^2+x-13)-1/1>=0}}}

Multiply top and bottom of {{{1/1}}} by LCD {{{(x^2+x-13)}}}

{{{(2x^2+6x-3)/(x^2+x-13)-(1*(x^2+x-13))/(1*(x^2+x-13))>=0}}}

{{{(2x^2+6x-3)/(x^2+x-13)-(x^2+x-13)/(x^2+x-13)>=0}}}

Combine the two numerators over the common denominator:

{{{((2x^2+6x-3)-(x^2+x-13))/(x^2+x-13)>=0}}}

Remove the parentheses:

{{{(2x^2+6x-3-x^2-x+13))/(x^2+x-13)>=0}}}

Combine like terms on top:

{{{(x^2+5x+10))/(x^2+x-13)>=0}}}

Find all critical values by setting numerator and
denominator = 0.

Setting numerator = 0

{{{x^2+5x+10=0}}}

discriminant = {{{b^2-4*a*c}}}

discriminant = {{{5^2-4*1*10=25-40=-15}}}

discriminant is negative, so we don't
get any critical values from the numerator.

Setting denominator = 0

{{{x^2+x-13=0}}}

discriminant = {{{b^2-4*a*c}}}

discriminant = {{{1^2-4*1*(-13)=1+52=53}}}

So we do get critical values from the
denominator:

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

{{{x = (-b +- sqrt(discriminant ))/(2*a) }}}

{{{x = (-1 +- sqrt(53))/(2*1) }}}

{{{x = (-1 +- sqrt(53))/2 }}}

So the two critical values are

{{{x = (-1 + sqrt(53))/2 }}} and {{{ (-1 - sqrt(53))/2 }}}

These are approximately 3.14 and -4.14

Put these on a number line:

------o----------------------o-----
-6 -5 -4 -3 -2 -1  0  1  2  3  4  5  

Choose a whole number value left of -4.14.
for a test value.  The easiest is -5

Substitute in 
{{{(x^2+5x+10)/(x^2+x-13)}}}
{{{((-5)^2+5(-5)+10)/((-5)^2+(-5)-13)}}}
{{{(25-25+10)/(25-5-13)}}}
{{{10/(25-5-13)}}}
{{{10/7}}}

That is a positive number, so we put + signs over 
that part of the number line:

 + + +
------o----------------------o-----
-6 -5 -4 -3 -2 -1  0  1  2  3  4  5

Next choose a whole number value between the
two critical points for a test value.  The 
easiest is 0

Substitute in 
{{{(x^2+5x+10)/(x^2+x-13)}}}
{{{((0)^2+5(0)+10)/((0)^2+(0)-13)}}}
{{{10/(-13)}}}
{{{10/(-13)}}}
{{{-10/13}}}

That is negative so we put - signs over that part 
of the number line:

 + + +  - - - - - - - - - - 
------o----------------------o-----
-6 -5 -4 -3 -2 -1  0  1  2  3  4  5

Choose a whole number value right of 3.14.
for a test value.  The easiest is 4

Substitute in 
{{{(x^2+5x+10)/(x^2+x-13)}}}
{{{((4)^2+5(4)+10)/((4)^2+(4)-13)}}}
{{{(16-20+10)/(16+4-13)}}}
{{{6/7}}}

That is a positive number, so we put + signs over 
that part of the number line:

 + + +  - - - - - - - - - -    + + +
------o----------------------o------
-6 -5 -4 -3 -2 -1  0  1  2  3  4  5

Since the inequality of 

{{{(x^2+5x+10))/(x^2+x-13)>=0}}}

is {{{"">=0}}}

we choose the intervals with + signs.

We cannot include the end points themselves

because they cause the denominator to be 0.

So the solution set is

({{{-infinity}}},{{{(-1 - sqrt(53))/2}}}) U ({{{(-1 + sqrt(53))/2 }}}, {{{infinity}}})

Edwin</pre>