Question 206101
I really need some help for this problems because I have been confuse of this problems since the start and now the Examination is coming near and this is going to be added, I really will appreciate some help in this questions:

---find the value of k so that the equation {{{ 3x^2-7x+k=0 }}} will have equal roots.
<pre><font size = 4 color = "indigo"><b>
Find the discriminant {{{b^2-4ac=(-7)^2-4(3)(k)=49-12k}}}

To have equal roots the discriminant must equal 0, so

{{{49-12k=0}}}
{{{-12k=-49}}}
{{{k=(-49)/(-12)}}}
{{{k=49/12}}}

To check, we substitute {{{49/12}}} for {{{k}}} in

{{{ 3x^2-7x+k=0 }}}
{{{ 3x^2-7x+49/12=0 }}}
Multiply through by 12 to clear of fractions:

{{{12*3x^2-12*7x+12*(49/12)=12*0}}}

{{{36x^2-84x+49=0}}}

{{{(6x-7)(6x-7)=0}}}

{{{6x-7=0}}}; {{{6x-7=0}}}
{{{6x=7}}}  ; {{{6x=7}}}
{{{x=7/6}}} ; {{{x=7/6}}}

So the roots are equal. 

----------------------------------- 

</b></pre></font>
---find the value of k so that the equation {{{ 3x^2-(5+k)x-17=0 }}} will have one roots numerically equal but opposite in sign
<pre><font size = 4 color = "indigo"><b>
Then it will have to be true that we also get zero when we 
substitute {{{-x}}} for {{{x}}}:

{{{ 3(-x)^2-(5+k)(-x)-17=0 }}}
{{{3x^2+(5+k)x-17=0}}}

And since we started with {{{ 3x^2-(5+k)x-17=0 }}}
We can set the left sides equal:

{{{3x^2+(5+k)x-17= 3x^2-(5+k)x-17=0 }}}

{{{3x^2+5x+kx-17= 3x^2-5x-kx-17=0 }}}

{{{10x+2kx=0 }}}

Divide thru by 2

{{{5x+kx=0}}}

Factor out x

{{{x(5+k)=0}}}

{{{x=0}}}; {{{5+k=0}}} 
           {{{k=-5}}}

So the value of k that will cause the
roots to be numerically equal but opposite
in sign.

Check by substituting -5 for k and solve:

{{{ 3x^2-(5+k)x-17=0 }}}
{{{ 3x^2-(5+(-5))x-17=0 }}}
{{{ 3x^2-17=0 }}}
{{{3x^2=17}}}
{{{x^2=17/3}}}
{{{x=""+-sqrt(17/3)=""+-sqrt((17*3)/(3*3))=""+-sqrt(51)/3}}}

{{{sqrt(51)/3}}} and {{{-sqrt(51)/3}}} are numerically 
equal but opposite in sign.

</b></pre></font>
--what should be the range of the value of k so that the equation 
{{{ 2x^2+3kx-9=0 }}} will have real and unequal roots.
<pre><font size = 4 color = "indigo"><b>
That is when the discriminant is positive

Find the discriminant {{{b^2-4ac=(3k)^2-4(2)(-9)=9k^2+72}}}

"Positive" means the same as "greater than 0"

So we set {{{9k^2+72>0}}}

Divide through by 9

{{{k^2+8>0}}}

Since {{{k^2}}} is never negative the left side will always
positive no matter what value of k we use, so all values of 
k will yield real and unequal roots.

So {{{-infinity<k<infinity}}}

</b></pre></font>
---what should be the value of k so that the equation {{{ x^2+(k-2)x+4=0 }}}
will have equal roots?
<pre><font size = 4 color = "indigo"><b>
Find the discriminant {{{b^2-4ac=(k-2)^2-4(1)(4)=4(k-2)^2-16}}}

To have equal roots the discriminant must equal 0, so

{{{4(k-2)^2-16=0}}}

Divide thru by 4

{{{(k-2)^2-4=0}}}

{{{(k-2)^2=4}}}

{{{k-2=""+-sqrt(4)}}}

{{{k-2=""+-2}}}

{{{k=2+-2}}}

Using the {{{"+"}}}, {{{k=2+2=4}}}

Using the {{{"-"}}}, {{{k=2-2=0}}}

Edwin</pre>