Lesson HOW TO solve equations containing Quadratic Terms under the Absolute Value sign. Lesson 2

Source code of 'HOW TO solve equations containing Quadratic Terms under the Absolute Value sign. Lesson 2'

This Lesson (HOW TO solve equations containing Quadratic Terms under the Absolute Value sign. Lesson 2) was created by by ikleyn(52776)

: View Source, Show
About ikleyn:

<H2>How to solve equations containing Quadratic Terms under the Absolute Value sign. Lesson 2</H2>

In this lesson you will see how to solve equations containing Quadratic Terms under the Absolute Value sign.
This is the starting lesson on this subject. We consider, one after the other, the solution procedure for the following examples:

a) &nbsp;{{{abs(x^2-8x+12)}}} = {{{x+4}}};

b) &nbsp;{{{abs(x^2-8x+12)}}} = {{{-2x+13}}};

c) &nbsp;{{{abs(x^2-8x+12)}}} = {{{x}}}.

The starting lesson on this subject is  
&nbsp;&nbsp;&nbsp;&nbsp;<A HREF=http://www.algebra.com/algebra/homework/absolute-value/HOW-TO-solve-equations-containing-Quadratic-Terms-under-Abs-Value-sign-L-1.lesson>How to solve equations containing Quadratic Terms under the Absolute Value sign. Lesson 1</A> 
where more simple cases are considered.

The goal is to get you acquainted with the different typical situations you can meet solving this kind of problems. 
When you learn how to solve these equations, you will easily proceed in many other cases.

<B>Problem 1</B>

Solve absolute value equation &nbsp;a) &nbsp;{{{abs(x^2-8x+12)}}} = {{{x+4}}}; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(1)

Solve absolute value equation &nbsp;b) &nbsp;{{{abs(x^2-8x+12)}}} = {{{-2x+13}}}; &nbsp;&nbsp;(2)

Solve absolute value equation &nbsp;c) &nbsp;{{{abs(x^2-8x+12)}}} = {{{x}}}. &nbsp;&nbsp;&nbsp;&nbsp&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(3)

<B>Solution to &nbsp;1a)</B>

<TABLE> 
  <TR>
  <TD>
By the definition

&nbsp;&nbsp;&nbsp;&nbsp;{{{abs(x^2-8x+12)}}} = &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;{{{x^2-8x+12}}}, &nbsp;&nbsp;if &nbsp;{{{x^2-8x+12}}} >= 0, &nbsp;and &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;  
&nbsp;&nbsp;&nbsp;&nbsp;{{{abs(x^2-8x+12)}}} = &nbsp;{{{-(x^2-8x+12)}}}, &nbsp;if &nbsp;{{{x^2-8x+12}}} < 0.

The equation &nbsp;{{{x^2-8x+12}}} = {{{0}}} &nbsp;has the roots &nbsp;{{{x=2}}} &nbsp;and&nbsp; {{{x=6}}}. 
Hence, we should try to solve the equation

{{{x^2-8x+12}}} = {{{x+4}}} &nbsp;over the ranges &nbsp;{{{x}}} <= {{{2}}} and &nbsp;{{{x}}} >= {{{6}}},&nbsp;&nbsp;&nbsp;&nbsp;(4)

where the quadratic function &nbsp;{{{x^2-8x+12}}}&nbsp; is positive, and the equation

{{{-(x^2-8x+12)}}} = {{{x+4}}} &nbsp;over the range &nbsp;{{{2}}} < {{{x}}} < {{{6}}}, &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(5)

where the quadratic function &nbsp;{{{x^2-8x+12}}}&nbsp; is negative.
 </TD>
  <TD> 
{{{graph( 250, 250, -1.2, 9.2, -6, 18,
          x^2-8x+12
)}}}

<B>Figure 1a</B>. Plot &nbsp;{{{y}}} = {{{x^2-8x+12}}}&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;

</TD>
  <TD> 
{{{graph( 250, 250, -1.2, 9.2, -6, 18,
          abs(x^2-8x+12),
          x+4
)}}}

<B>Figure 1a'</B>. Plot &nbsp;{{{y}}} = {{{abs(x^2-8x+12)}}} 
and the linear function &nbsp;{{{y}}} = {{{x+4}}}
  </TD>
  </TR>
</TABLE>
Simplifying the equation &nbsp;(4)&nbsp; you get &nbsp;{{{x^2-9x+8=0}}}. &nbsp;The solutions are &nbsp;{{{x[1]=1}}} &nbsp;and&nbsp; {{{x[2]=8}}}. &nbsp;They are in the ranges &nbsp;{{{x[1]}}}<={{{2}}} &nbsp;and&nbsp; {{{x[2]}}}>={{{6}}}, &nbsp;hence, 
satisfy all the requirements. 
Simplifying the equation &nbsp;(5)&nbsp; you get &nbsp;{{{x^2-7x+16=0}}}, &nbsp;which has no real solutions.

Thus, &nbsp;the equation &nbsp;(1)&nbsp; has two solutions &nbsp;{{{x=1}}} &nbsp;and&nbsp; {{{x=8}}}.

In &nbsp;<B>Figure 1a'</B>&nbsp; the red line represents the plot of the function &nbsp;{{{y}}} = {{{abs(x^2-8x+12)}}}. &nbsp;The green line represents the linear function {{{y}}} = {{{x+4}}}. 
The intersection points represent the two solutions we just found.

<B>Solution to &nbsp;1b)</B>

<TABLE> 
  <TR>
  <TD>
By the definition

The equation &nbsp;{{{x^2-8x+12}}} = {{{0}}} &nbsp;has the roots &nbsp;{{{x=2}}} &nbsp;and&nbsp; {{{x=6}}}. 
Hence, we should try to solve the equation

{{{x^2-8x+12}}} = {{{-2x+13}}} &nbsp;over the ranges &nbsp;{{{x}}} <= {{{2}}} and &nbsp;{{{x}}} >= {{{6}}},&nbsp;&nbsp;&nbsp;&nbsp;(6)

where the quadratic function &nbsp;{{{x^2-8x+12}}}&nbsp; is positive, and the equation

{{{-(x^2-8x+12)}}} = {{{-2x+13}}} &nbsp;over the range &nbsp;{{{2}}} < {{{x}}} < {{{6}}}, &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(7)

where the quadratic function &nbsp;{{{x^2-8x+12}}}&nbsp; is negative.
 </TD>
  <TD> 
{{{graph( 250, 250, -1.2, 9.2, -6, 18,
          abs(x^2-8x+12),
          -2x+13.15
)}}}

<B>Figure 1b</B>. Plot &nbsp;{{{y}}} = {{{abs(x^2-8x+12)}}} 
and the linear function &nbsp;{{{y}}} = {{{-2x+13}}}
  </TD>
  </TR>
</TABLE>
Simplifying the equation &nbsp;(6)&nbsp; you get &nbsp;{{{x^2-6x-1=0}}}. &nbsp;The solutions are &nbsp;{{{x[1]=3-sqrt(10)}}}=~{{{-0.162}}} &nbsp;and&nbsp; {{{x[2]=3+sqrt(10)}}}=~{{{6.162}}}. 
They are in the ranges &nbsp;{{{x[1]}}}<={{{2}}} &nbsp;and&nbsp; {{{x[2]}}}>={{{6}}}, &nbsp;hence, satisfy all the requirements. 
Simplifying the equation &nbsp;(7)&nbsp; you get &nbsp;{{{x^2-10x+25=0}}}. &nbsp;The solutions are &nbsp;{{{x[1]=5}}} &nbsp;and&nbsp; {{{x[2]=5}}}. &nbsp;(Two solutions merge into one). &nbsp;The solutions are 
in the range &nbsp;{{{2}}} < {{{x}}} < {{{6}}}, &nbsp;hence, satisfy all the requirements.

Thus, &nbsp;the equation &nbsp;(2)&nbsp; has three solutions &nbsp;{{{x=3-sqrt(10)}}}, &nbsp; {{{x=3+sqrt(10)}}} &nbsp;and &nbsp;{{{x=5}}}.

In &nbsp;<B>Figure 1b</B>&nbsp; the red line represents the plot of the function &nbsp;{{{y}}} = {{{abs(x^2-8x+12)}}}. &nbsp;The green line represents the linear function {{{y}}} = {{{-2x+13}}}. 
The intersection points represent the three solutions we just found.

<B>Solution to &nbsp;1c)</B>

<TABLE> 
  <TR>
  <TD>
Again, by the definition

The equation &nbsp;{{{x^2-8x+12}}} = {{{0}}} &nbsp;has the roots &nbsp;{{{x=2}}} &nbsp;and&nbsp; {{{x=6}}}. 
Hence, we should try to solve the equation

{{{x^2-8x+12}}} = {{{x}}} &nbsp;over the ranges &nbsp;{{{x}}} <= {{{2}}} and &nbsp;{{{x}}} >= {{{6}}},&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(8)

where the quadratic function &nbsp;{{{x^2-8x+12}}}&nbsp; is positive, and the equation

{{{-(x^2-8x+12)}}} = {{{x}}} &nbsp;over the range &nbsp;{{{2}}} < {{{x}}} < {{{6}}}, &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(9)

where the quadratic function &nbsp;{{{x^2-8x+12}}}&nbsp; is negative.
 </TD>
  <TD> 
{{{graph( 250, 250, -1.2, 9.2, -6, 18,
          abs(x^2-8x+12),
          x
)}}}

<B>Figure 1c</B>. Plot &nbsp;{{{y}}} = {{{abs(x^2-8x+12)}}} 
and the linear function &nbsp;{{{y}}} = {{{x}}}
  </TD>
  </TR>
</TABLE>
Simplifying the equation &nbsp;(8)&nbsp; you get &nbsp;{{{x^2-9x+12=0}}}. &nbsp;The solutions are &nbsp;{{{x[1]=(9-sqrt(33))/2}}}=~{{{1.628}}} &nbsp;and&nbsp; {{{x[2]=(9+sqrt(33))/2}}}=~{{{7.372}}}. 
They are in the ranges &nbsp;{{{x[1]}}}<={{{2}}} &nbsp;and&nbsp; {{{x[2]}}}>={{{6}}}, &nbsp;hence, satisfy all the requirements. 
Simplifying the equation &nbsp;(9)&nbsp; you get &nbsp;{{{x^2-7x+12=0}}}. &nbsp;The solutions are &nbsp;{{{x[1]=3}}} &nbsp;and&nbsp; {{{x[2]=4}}}. &nbsp;They are in the range &nbsp;{{{2}}} < {{{x}}} < {{{6}}}, &nbsp;hence, 
satisfy all the requirements.

Thus, &nbsp;the equation &nbsp;(3)&nbsp; has four solutions &nbsp;{{{x=(9-sqrt(33))/2}}}, &nbsp; {{{x=(9+sqrt(33))/2}}}, &nbsp;{{{x=3}}} &nbsp;and &nbsp;{{{x=4}}}.

In &nbsp;<B>Figure 1c</B>&nbsp; the red line represents the plot of the function &nbsp;{{{y}}} = {{{abs(x^2-8x+12)}}}. &nbsp;The green line represents the constant function {{{y}}} = {{{x}}}. 
The intersection points represent the four solutions we just found.

Below the plots of the &nbsp;<B>Figure 1a</B> - <B>Figure 1c</B>&nbsp; are presented again for your convenience.

<TABLE> 
  <TR>
  <TD> 
{{{graph( 250, 250, -1.2, 9.2, -6, 18,
          abs(x^2-8x+12),
          x+4
)}}}

<B>Figure 1a</B>. Plot &nbsp;{{{y}}} = {{{abs(x^2-8x+12)}}}&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 
and the linear function &nbsp;{{{y}}} = {{{x+4}}}
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(two roots)
  </TD>
   <TD>
{{{graph( 250, 250, -1.2, 9.2, -6, 18,
          abs(x^2-8x+12),
          -2x+13.15
)}}}

<B>Figure 1b</B>. Plot &nbsp;{{{y}}} = {{{abs(x^2-8x+12)}}}&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 
and the linear function &nbsp;{{{y}}} = {{{-2x+13}}}
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(three roots)
  </TD>
   <TD>
{{{graph( 250, 250, -1.2, 9.2, -6, 18,
          abs(x^2-8x+12),
          x
)}}}

<B>Figure 1c</B>. Plot &nbsp;{{{y}}} = {{{abs(x^2-8x+12)}}} 
and the linear function &nbsp;{{{y}}} = {{{x}}}
&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(four roots)
  </TD>
  </TR>
</TABLE>

Note that the solution strategy is the same in all examples. &nbsp;It is to break up the entire set of real numbers into sub-domains (ranges) 
where the absolute value of a quadratic term is a quadratic function, &nbsp;and then to solve the corresponding quadratic equations in each sub-domain (range).

My other lessons on &nbsp;<B>Absolute Value equations</B>&nbsp; in this site are

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/absolute-value/-Absolute-Value-equations.lesson>Absolute Value equations</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/absolute-value/HOW-TO-solve-equations-containing-Linear-Terms-under-Abs-Value-sign-L-1.lesson>HOW TO solve equations containing Linear Terms under the Absolute Value sign. Lesson 1</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/absolute-value/HOW-TO-solve-equations-containing-Linear-Terms-under-Abs-Value-sign-L-2.lesson>HOW TO solve equations containing Linear Terms under the Absolute Value sign. Lesson 2</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/absolute-value/HOW-TO-solve-equations-containing-Linear-Terms-under-Abs-Value-sign-L-3.lesson>HOW TO solve equations containing Linear Terms under the Absolute Value sign. Lesson 3</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/absolute-value/HOW-TO-solve-eqns-containing-Lin-Terms-under-Abs-value-L4.lesson>HOW TO solve equations containing Linear Terms under the Absolute Value sign. Lesson 4</A>

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/absolute-value/HOW-TO-solve-equations-containing-Quadratic-Terms-under-Abs-Value-sign-L-1.lesson>HOW TO solve equations containing Quadratic Terms under the Absolute Value sign. Lesson 1</A>

See also the lesson &nbsp;<A HREF=http://www.algebra.com/algebra/homework/absolute-value/Review-of-lessons-on-Absolute-Value-equations.lesson>OVERVIEW of lessons on Absolute Value equations</A>&nbsp; under the current topic 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.