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# algebra 2 help

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Q1)

Q2)

Q3)

Oct 30, 2018

#1
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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
+316
+2

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