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Solve the equation `sqrt(x)=x-2` |
Answer» We have `sqrt(x)=x-2` On squaring both sides we obtain `x=(x-2)^(2)` `impliesx^(2)-5x+4=0implies(x-1)(x-4)=0` `:.x_(1)=1` and `x_(2)=4` Hence `x_(1)=4` satisfies the original equation, but `x_(2)=1` does not satisfy the original equation. `:.x_(2)=1` is the extraneous root. |
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