Question 202022
Could somebody guide me about this which is about the relation between the roots and the coefficients?

For what values of m will the equation {{{x^2-4x+7=m(x-1)}}} have
a) one root the reciprocal of the other,
b) one root equal zero
c) equal roots
<pre><font size = 4 color = "indigo"><b>
To do any of those we have to first get

{{{x^2-4x+7=m(x-1)}}} in the form {{{x^2+Ax+B=0}}}
{{{x^2-4x+7=mx-m}}}
{{{x^2-4x-mx+7+m=0}}}
{{{x^2-(4+m)x+(7+m)=0}}}

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</pre></font></b>
a) one root the reciprocal of the other
<pre><font size = 4 color = "indigo"><b>
Suppose one root is r and the other is {{{1/r}}}, then
the quadratic equation with this property and leading
coefficient 1 is found this way:

{{{(x-r)(x-1/r)=0}}}

{{{x^2-(r+1/r)x+1=0}}}

Therefore this must be the same equation as 

{{{x^2-(4+m)x+(7+m)=0}}}

So we equate like parts:

{{{system(-(r+1/r)=-(4+m),1=(7+m))}}}

or simplifying:

{{{system(r+1/r=4+m,-6=m)}}}

To check that, we substitute {{{-6}}} 
for {{{m}}}:

{{{x^2-4x+7=m(x-1)}}}
{{{x^2-4x+7=-6(x-1)}}}
{{{x^2-4x+7=-6x+6}}}
{{{x^2+2x+1=0}}}
{{{(x+1)(x+1)=0}}}
{{{x+1=0}}}, {{{x+1=0}}}    
{{{x=-1}}}  , {{{x=-1}}}

So both roots are equal and -1, but -1 is the
reciprocal of -1.

So the answer to (a) is {{{m=-6}}}
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</pre></font></b>
b) one root equal zero
<pre><font size = 4 color = "indigo"><b>
Suppose one root is 0 and the other is r, then
the quadratic equation with this property and leading
coefficient 1 is found this way:

{{{(x-0)(x-r)=0}}}
{{{x(x-r)=0}}}
{{{x^2-rx=0}}}

Therefore this must be the same equation as 

{{{x^2-(4+m)x+(7+m)=0}}}

So we equate like parts:

{{{system(-r=-(4+m),0=(7+m))}}}

Simplifying:

{{{system(r=4+m,-7=m)}}}

So we see that m must be -7.

To check that, we substitute {{{-7}}} 
for {{{m}}}:

{{{x^2-4x+7=m(x-1)}}}
{{{x^2-4x+7=-7(x-1)}}}
{{{x^2-4x+7=-7x+7}}}
{{{x^2+3x=0}}}
{{{x(x+3)=0}}}
{{{x=0}}}, {{{x+3=0}}}    
           {{{x=-3}}}

One root is 0, so we are correct.

So the answer to (b) is {{{m=-7}}}

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</pre></font></b>
c) equal roots
<pre><font size = 4 color = "indigo"><b>
Suppose one root is r and the other is also r, then
the quadratic equation with this property and leading
coefficient 1 is found this way:

{{{(x-r)(x-r)=0}}}
{{{(x-r)(x-r)=0}}}
{{{x^2-2rx+r^2=0}}}

Therefore this must be the same equation as 

{{{x^2-(4+m)x+(7+m)=0}}}

So we equate like parts:

{{{system(-2r=-(4+m),r^2=(7+m))}}}

Simplifying:

{{{system(2r=4+m,r^2=7+m)}}}

Solve both equations for m

{{{system(m=2r-4,m=r^2-7)}}}

Equate the right sides since both equal m:

{{{2r-4=r^2-7}}}
{{{-r^2+2r+3=0}}}
{{{r^2-2r-3=0}}}
{{{(r+1)(r-3)=0}}}
{{{r+1=0}}}, {{{r-3=0}}}    
{{{r=-1}}} , {{{r=3}}}

Substituting {{{r=-1}}} into

{{{m=2r-4}}}
{{{m=2(-1)-4}}}
{{{m=-2-4}}}
{{{m=-6}}}

Substituting {{{r=3}}} into

{{{m=2r-4}}}
{{{m=2(3)-4}}}
{{{m=6-4}}}
{{{m=2}}}

We don't need to check m=-6 for that was
the value of m in part (a), and we knew 
that in that case the roots were not only 
reciprocals but they were also equal.

Checking m=2

{{{x^2-4x+7=m(x-1)}}}
{{{x^2-4x+7=2(x-1)}}}
{{{x^2-4x+7=2x-2}}}
{{{x^2-6x+9=0}}}
{{{(x-3)(x-3)=0}}}
{{{x-3=0}}}, {{{x-3=0}}}    
{{{x=3}}}  , {{{x=3}}}

So there are two answers to (c),

{{{m=-6}}} and {{{m=2}}}.

Edwin</pre>