Question 162002
pls.help me solve the following quadratic equation by factoring:
<pre><font size = 4 color = "indigo"><b>
{{{2x^2-3x+1=0}}}

Multiply the {{{2}}} in front by the {{{matrix(1,1,"+1")}}},
getting {{{2}}}.  So we write down

{{{1*2}}}

as the only way to find two factors of {{{2}}}.

Next look at the sign of the last term {{{matrix(1,1,"+1")}}} 
on the left.  It is {{{matrix(1,1,"+")}}} so beside {{{1*2}}}
we write the sum {{{1+2}}}, and what it equals, {{{3}}}:.  

{{{1*2}}}   {{{1+2=3}}}

We observe that the absolute
value of the coefficient of the middle term on the left of
{{{2x^2-3x+1=0}}} is {{{3}}} so we replace the {{{3}}} by
{{{(1+2)}}}, so

{{{2x^2-3x+1=0}}}   becomes

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

Now place the {{{x}}} on the left of the {{{(1+2)}}} as

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

Us the distributive property to multiply out {{{-x(1+2)}}}
as {{{-x-2x}}}. So now we have:

{{{2x^2-x-2x+1=0}}}

Out of the first two terms on the left ONLY, {{{2x^2-x}}},
we can factor out {{{x}}}, getting {{{x(2x-1)}}} 

Out of the last two terms on the left ONLY, {{{-2x+1}}},
we can factor out {{{-1}}}, getting {{{-1(2x-1)}}}.

[Notice when the first of the last two terms begins
with a "-" sign, always take out a "-" term, which
will invariably change the sign of the second term.]

So we have

{{{x(2x-1)-1(2x-1)=0}}}

Now we take out the parentheses {{{(2x-1)}}} as a
common factor and leave the {{{x}}} and the {{{-1}}}
to put inside parentheses, like this:

{{{(2x-1)(x-1)=0}}}

Noqw we use the zero-factor principle where we 
set each factor = 0:

{{{matrix(3,3, 2x-1=0, "," , x-1=0, 2x=1, ",",  x=1,x=1/2," "," ")}}}
  
So the solution set is {{{matrix(1,5, "{", 1/2, ",", 1, "}")}}}

That's just the first one.  If I have time I will come back
and finish the rest.  So check this site later, for I will
probably do some more.

Edwin</pre></font></b> 
{{{4x^2+9x=9}}}
{{{2y^2+y=-11y}}}
{{{y^2-7-44=0}}}
{{{6x^2+11x-1=0}}}