Lesson Advanced problems on solving equations containing radicals

Source code of 'Advanced problems on solving equations containing radicals'

This Lesson (Advanced problems on solving equations containing radicals) was created by by ikleyn(52777)

: View Source, Show
About ikleyn:

<H2>Advanced problems on solving equations containing radicals</H2>

<H3>Problem 1</H3>Solve an equation  &nbsp;{{{sqrt(x)}}} + {{{sqrt(1-x)}}} + {{{sqrt(x*(1-x))}}} = 1

<B>Solution</B>

<pre>
{{{sqrt(x)}}} + {{{sqrt(1-x)}}} + {{{sqrt(x*(1-x))}}} = 1.      (1)

The domain, where all included functions are defined, is the segment  [0,1].

Two obvious solutions to the given equation in this domain are x= 0  and  x= 1.

Below I will show that the given equation <U>HAS NO other solutions</U>.

Indeed, let  0 < x < 1.

Then        {{{sqrt(x)}}} is defined and is positive number  {{{sqrt(x)}}} > 0.

Similarly,  {{{sqrt(1-x)}}} is defined and is positive number  {{{sqrt(1-x)}}} > 0.

For any two real positive numbers "a" and "b" the following inequality is valid

a + b > {{{sqrt(a^2 + b^2)}}}.    (2)

To prove it, square both sides. You will get

a^2 + 2ab + b^2 > a^2 + b^2,

which is valid for all positive "a" and "b".

Now apply the inequality (2) for  a= {{{sqrt(x)}}}  and  b= {{{sqrt(1-x)}}}.  You will get

{{{sqrt(x)}}} + {{{sqrt(1-x)}}} > {{{sqrt((sqrt(x))^2 + (sqrt(1-x))^2)}}} = {{{sqrt(x + 1-x)}}} = {{{sqrt(1)}}} = 1.

Thus,  the sum  {{{sqrt(x)}}} + {{{sqrt(1-x)}}}  at  0 < x < 1  is just greater than 1.

With the added positive addend  {{{sqrt(x*(1-x))}}},  the sum  {{{sqrt(x)}}} + {{{sqrt(1-x)}}} + {{{sqrt(x*(1-x))}}}  is just even more than 1.

Therefore,  the sum  {{{sqrt(x)}}} + {{{sqrt(1-x)}}} + {{{sqrt(x*(1-x))}}}  can not be equal to 1  at  0 < x < 1.

Thus, it  <U>PROVED</U>  that the given equation has no solutions inside the segment [0,1].

So, the endpoints  x= 0  and  x= 1 are the only solutions.
</pre>

<H3>Problem 2</H3>Solve an equation  &nbsp;{{{sqrt(10-x)}}} + {{{sqrt(3+x)}}} + {{{2*sqrt(30+7x-x^2)}}} = 17.

<B>Solution</B>

<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>

<H3>Problem 3</H3>Solve an equation &nbsp;&nbsp;{{{(7 + sqrt(x))^(1/3)}}} + {{{(7 - sqrt(x))^(1/3)}}} = 2.

<B>Solution</B>

<pre>
Your starting equation is

{{{(7+sqrt(x))^(1/3)}}} + {{{(7-sqrt(x))^(1/3)}}} = 2.    (1)

Let  a = {{{(7+sqrt(x))^(1/3)}}};  b = {{{(7-sqrt(x))^(1/3)}}}.

Raise both sides of equation (1) to degree 3.

Use

{{{(a + b)^3}}} = {{{a^3 + b^3}}} + 3ab*(a+b),

{{{a^3 + b^3}}} = {{{(7+sqrt(x))}}}  + {{{(7-sqrt(x))}}} = 14,

a + b = 2   <<<---===  given by equation,

ab = {{{(49 - x)^(1/3)}}}.

You will get

{{{14 + 3*(49-x)^(1/3)*2}}} = 8.

Simplify and find x

{{{(49 - x)^(1/3)}}} = {{{(8-14)/6}}},

{{{(49 - x)^(1/3)}}} = -1.

Raise both sides of the last equation to degree 3

49 - x = -1,

49 + 1 = x,

x     = 50.

<U>ANSWER</U>.  x = 50.

<U>CHECK</U>.   {{{(7 + sqrt(50))^(1/3)}}} + {{{(7 - sqrt(50))^(1/3)}}} = 2.414213562 + (-0.414213562) = 2.0000000
</pre>

My other lesson on solving equations containing radicals is

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/Radicals/Solving-equations-containing-radicals.lesson>HOW TO solve equations containing radicals</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/Radicals/Solving-systems-of-equations-containing-radicals.lesson>Solving systems of equations containing radicals</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/Radicals/OVERVIEW-of-my-lessons-on-solving-equations-containing-radicals.lesson>OVERVIEW of my lessons on solving equations containing radicals</A>

in this site.

Use this file/link &nbsp;<A HREF=https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-I - YOUR ONLINE TEXTBOOK</A>&nbsp; to navigate over all topics and lessons of the online textbook &nbsp;ALGEBRA-I.