Question 1152365
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<h3>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;The solution can be done in much simpler way.</H3>


<pre>
{{{sqrt(10-x)}}} + {{{sqrt(3+x)}}} + {{{2*sqrt(30+7x-x^2)}}} = 17      (1)

{{{sqrt(10-x)}}} + {{{sqrt(3+x)}}} = 17 - {{{2*sqrt(30+7x-x^2)}}} 


Square both sides


(10-x) + {{{2*sqrt(10-x)*sqrt(3+x)}}} + (3+x) = 17 - {{{68*sqrt(30+7x-x^2)}}} + 4*(30 +7x - x^2)


Notice that  (10-x)*(3+x) = 30 + 7x - x^2, and continue transform preceding equations


(10-x) + {{{2*sqrt(30+7x-x^2)}}} + (3+x) = 289 - {{{68*sqrt(30+7x-x^2)}}} + 4*(30 +7x - x^2)

13 + {{{2*sqrt(30+7x-x^2)}}} = 289 - {{{68*sqrt(30+7x-x^2)}}} + 4*(30+7x-x^2)

0 = 276 - {{{70*sqrt(30+7x-x^2)}}} + 4*(30+7x-x^2)      (2)


Introduce new variable t = {{{sqrt(30+7x-x^2)}}}.


Then equation (2) takes the form


4t^2 - 70t + 276 = 0.


Solve it using the quadratic formula


{{{t{1,2]}}} = {{{(70 +- sqrt(70^2 - 4*4*276))/(2*4)}}} = {{{(-70 +- sqrt(484))/8}}} = {{{(-70 +- 22)/8}}}.


Case 1.  t = {{{(-70 + 22)/8}}} = -6.


         Then t = {{{sqrt(30+7x-x^2)}}} = -6 implies (after squaring both sides)

              30 + 7x - x^2 = 36

              x^2 - 7x + 6 = 0

              (x-1)*(x-6) = 0

              The roots are  x= 1  and  x= 6.

              You can easily check that both these roots satisfy the original equation.



Case 2.  t = {{{(-70 - 22)/8}}} = -11.5.


         Then t = {{{sqrt(30+7x-x^2)}}} = -11.5 implies (after squaring both sides)

              30 + 7x - x^2 = 132.25

              x^2 - 7x + 102.25 = 0

              Discriminant d = b^2 - 4ac = 7^2 - 4*102.25 is negative,

              Hence, this case does not produce real solutions.



The solution is completed.


The  <U>ANSWER</U>  is:  the original equation has two solutions  x= 1  and  x= 6.
</pre>

Solved.


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In her solution, tutor @MathLover1 makes a HUGE amount unnecessary work.


Also, on the way, she applies (without acknowledgement) Internet software to factor polynomials of high degrees (up to 8-th degree).


<pre>
    It is <U>IMPOSSIBLE</U> to get such factoring "by hands", without using software tools.
</pre>

Obviously, this way is not accessible for real school Math student, and it is NOT the WAY to solve such problem for educational purposes.


The way which I propose in my post, is a real way, and, actually, the <U>only real, accessible and right way to solve this problem</U>.