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Neil said that – 3 and 6 are the solutions to $square root of 3 x plus 18 end root equals x$ . What error did Neil make?

hint Hint:

Rearrange the equation and then solve for x.

The correct answer is: x = -3

Complete step by step solution:
Here we have the equation $square root of 3 x plus 18 end root equals x$
On squaring both the sides, we have $3 x plus 18 equals x squared$
On rearranging, we have $x squared minus 3 x minus 18 equals 0$
Now, this is a quadratic equation with a = 1, b = -3, c = -18
Roots can be found with, $x equals fraction numerator negative b plus-or-minus square root of b squared minus 4 a c end root over denominator 2 a end fraction$
$x equals fraction numerator 3 plus-or-minus square root of 9 minus 4 cross times 1 cross times negative 18 end root over denominator 2 cross times 1 end fraction$
On solving, we get x = 6, -3
Here, on substituting x = 6 or -3 in given equation we get
$square root of 3 cross times 6 plus 18 end root equals 6$
$not stretchy rightwards double arrow square root of 18 plus 18 end root equals square root of 36 equals 6$
Hence x = 6 works.
Now, $square root of 3 cross times negative 3 plus 18 end root equals negative 3$
$not stretchy rightwards double arrow square root of negative 9 plus 18 end root equals square root of 9 not equal to negative 3$
Hence x = -3 is an extraneous solution.
The error is that, one of the solutions, x = -3 is an extraneous solution.

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Neil said that – 3 and 6 are the solutions to . What error did Neil make?

Rearrange the equation and then solve for x.

The correct answer is: x = -3

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