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Q1)

Q2)

Q3)

 Oct 30, 2018

Best Answer 

 #1
avatar+316 
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A1) We want \(g^2-2g-24≠ 0\) so we will find when \(g^2-2g-24= 0\) so  \(g = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

\(g1=\frac{2+\sqrt{2^2-4(1)(-24)}}{2(1)}\)<=>\(g1=\frac{2+\sqrt{100}}{2}\)=\(\frac{2+10}{2}=\frac{12}{2}=6\) and

\(g2=\frac{2-\sqrt{2^2-4(1)(-24)}}{2(1)}\)<=>\(g2=\frac{2-\sqrt{100}}{2}=\frac{2-10}{2}=\frac{-8}{2}=-4\)

Finally \(g1=6,g2=-4\) so We want \(g^2-2g-24≠ 0\) so \(g≠ 6,g≠ -4\)

 

A2) \(\frac{(x+9)(x-2)}{(x-2)(2+x)}=\frac{(x+9)}{(2+x)}\times\frac{(x-2)}{(x-2)}=\frac{(x+9)}{(2+x)}\times1=\frac{(x+9)}{(2+x)}\)Correct the 3rd answer BUT we assume that the equation is well defined and \(x≠ 2\)

 

A3)\(\frac{9-x^2}{9x+27}=\frac{(3+x)(3-x)}{(9)(x+3)}=\frac{(x+3)(3-x)}{(9)(x+3)}=\frac{3-x}{9}\times\frac{(x+3)}{(x+3)}=\frac{3-x}{9}\times1=\frac{3-x}{9}\) BUT  like before we assume that the equation is well defined and \(x≠3\)

 

Hope this helps!

 Oct 30, 2018
 #1
avatar+316 
+2
Best Answer

A1) We want \(g^2-2g-24≠ 0\) so we will find when \(g^2-2g-24= 0\) so  \(g = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

\(g1=\frac{2+\sqrt{2^2-4(1)(-24)}}{2(1)}\)<=>\(g1=\frac{2+\sqrt{100}}{2}\)=\(\frac{2+10}{2}=\frac{12}{2}=6\) and

\(g2=\frac{2-\sqrt{2^2-4(1)(-24)}}{2(1)}\)<=>\(g2=\frac{2-\sqrt{100}}{2}=\frac{2-10}{2}=\frac{-8}{2}=-4\)

Finally \(g1=6,g2=-4\) so We want \(g^2-2g-24≠ 0\) so \(g≠ 6,g≠ -4\)

 

A2) \(\frac{(x+9)(x-2)}{(x-2)(2+x)}=\frac{(x+9)}{(2+x)}\times\frac{(x-2)}{(x-2)}=\frac{(x+9)}{(2+x)}\times1=\frac{(x+9)}{(2+x)}\)Correct the 3rd answer BUT we assume that the equation is well defined and \(x≠ 2\)

 

A3)\(\frac{9-x^2}{9x+27}=\frac{(3+x)(3-x)}{(9)(x+3)}=\frac{(x+3)(3-x)}{(9)(x+3)}=\frac{3-x}{9}\times\frac{(x+3)}{(x+3)}=\frac{3-x}{9}\times1=\frac{3-x}{9}\) BUT  like before we assume that the equation is well defined and \(x≠3\)

 

Hope this helps!

Dimitristhym Oct 30, 2018

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