Question 906578
{{{3sqrt(4x-7) + 2sqrt(9-x)= 17}}} 
<pre>
Let one of the square roots equal to a letter

Let {{{sqrt(9-x)=A}}}

Then we have:

{{{3sqrt(4x-7) + 2A= 17}}}

Isolate the remaining radical term on the left

{{{3sqrt(4x-7) = 17-2A}}}

Square both sides:

{{{(3sqrt(4x-7))^2 = (17-2A)^2}}}

{{{9(4x-7)=(17-2A)(17-2A)}}}

{{{36x-63=289-34A-34A+4A^2}}}

{{{36x-63=289-68A+4A^2}}}

{{{36x=352-68A+4A^2}}}

Now since  {{{A=sqrt(9-x)}}}, {{{A^2=9-x}}}

{{{36x=352-68sqrt(9-x)+4(9-x)}}}

{{{36x=352-68sqrt(9-x)+36-4x}}}

{{{36x=388-68sqrt(9-x)-4x}}}

Isolate the square root term:

{{{68sqrt(9-x)=388-40x}}}

Divide through by 4

{{{17sqrt(9-x)=97-10x}}}

Square both sides:

{{{(17sqrt(9-x))^2=(97-10x)^2}}}

{{{289(9-x)=(97-10x)(97-10x)}}}

{{{2601-289x=9409-970x-970x+100x^2}}}

{{{2601-289x=9409-1940x+100x^2}}}

{{{0=6808-1651x+100x^2}}}

{{{100x^2-1651x+6808=0}}}

That may factor, but it would take longer to
factor it than to use the quadratic formula,
so I'll just use the quadratic formula:

 {{{x = (-b +- sqrt( b^2-4ac ))/(2a) }}}
 
 {{{x = (-(-1651) +- sqrt((-1651)^2-4(100)(6808) ))/(2(100)) }}}

 {{{x = (1651 +- sqrt(2601) ))/200 }}}

 {{{x = (1651 +- 51 )/200 }}} 

Using the +

 {{{x = (1651 + 51 )/200 }}}

 {{{x = 1702/200 }}} 

 {{{x = 851/100}}}

Using the -

 {{{x = (1651 - 51 )/200 }}}

 {{{x = 1600/200 }}} 

 {{{x = 8}}}

So the solutions are {{{851/100}}} and {{{8}}}

But we must check to see if they are really solutions:

checking {{{851/100}}}

{{{3sqrt(4x-7) + 2sqrt(9-x)= 17}}}
{{{3sqrt(4(851/100)-7) + 2sqrt(9-851/100)= 17}}}
{{{3sqrt(851/25-7) + 2sqrt(9-851/100)= 17}}}
{{{3sqrt(851/25-175/25) + 2sqrt(900/100-851/100)= 17}}}
{{{3sqrt(676/25) + 2sqrt(49/100)= 17}}}
{{{3(26/5)+2(7/10)=17}}}
{{{78/5+14/10=17}}}
{{{156/10+14/10=17}}}
{{{170/10=17}}}
{{{17=17}}}

checking {{{8}}}

{{{3sqrt(4x-7) + 2sqrt(9-x)= 17}}}

{{{3sqrt(4*8-7) + 2sqrt(9-8)= 17}}}

{{{3sqrt(32-7)+2sqrt(1)=17}}}

{{{3sqrt(25)+2(1)=17}}}

{{{3*5+2=17}}}

{{{15+2-17}}}
{{{17=17}}}

They both check.

Edwin</pre>




Edwin</pre>