How do you solve #3-13s=3-8s+s^2#?

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Answer 1 · 2016-04-11T09:17:10

#s=-5#

Our aim is to isolate and find the value of #s#

#color(blue)(3-13s=3-8s+s^2#

Subtract #3# both sides

#rarrcolor(blue)(3)-13s-color(blue)(3)=color(blue)(3)-8s+s^2-color(blue)(3)#

#rarr-13s=-8s+s^2#

Add #8s# both sides

#rarrcolor(blue)(-13s)+color(blue)(8s)=color(blue)(-8s)+s^2+color(blue)(-8)#

#rarr-5s=s^2#

From now use logic!

We know that #color(brown)(s^2=s*s#

So,

#rarr-5s=s*s#

Divide both sides by #s#

#rarr(-5cancelcolor(blue)(s))/cancelcolor(blue)(s)=(s*cancelcolor(blue)(s))/cancelcolor(blue)(s)#

#color(green)(rArr-5=s#