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Find the non-extraneous solutions of...
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\[\sqrt{x+9}-5=x+4\]
@perl
@Jhannybean
@Secret-Ninja
@Abhisar
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Well for a start, you can add the 5 to the other side. Square root x+9=x+9 Square each of them x+9=x^2+18x+81 Minus the first term. x^2+17x+72=0 Factor that to (x+9)(x+8) You tell me what is extraneous.
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