Question 176792
{{{sqrt(r+4)+2=r}}}
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{{{sqrt(r+4)=r-2}}} subtracted 2 from both sides
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{{{sqrt(r+4)^2=(r-2)^2}}}square both sides
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{{{r+4=r^2-4r+4}}}carried out operations
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{{{r^2-5r=0}}}combined terms all on one side
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{{{r(r-5)=0}}}
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r=0 and r=5
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at times one will get extraneous roots when a term is squared
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r=0 is one such root
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therefore the only solution is r=5