How do you solve and check for extraneous solutions in $\sqrt{5 x} = - 5$ $sqrt(5x) = -5$ ?

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How do you solve and check for extraneous solutions in $\sqrt{5 x} = - 5$ $sqrt(5x) = -5$ ?

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Answer 1 · 2015-07-31T20:41:05

With the usual meaning for $\sqrt{5 x}$ $sqrt(5x)$ as the principal square root, this equation has no solutions.

Explanation:

The usual definition given is that
$y = \sqrt{x}$ $y = sqrtx$ if and only if (both $y^{2} = x$ $y^2 = x$ and $y \geq 0$ $y >= 0$ )

This equation asks us, to find a number whose non-negative square root is $- 5$ $-5$ . There is no such number.

(This question is similar to asking us to solve $| 2 x | = - 10$ $abs(2x) = -10$ . It is built into the definition of the expression on the left that it is a non-negative number.)

But
If the point of the exercise is to investigate extraneous solutions, then:

We might try squaring both sides (to get rid of the square root on the left)

$\sqrt{5 x} = - 5$ $sqrt(5x) = -5$

${(\sqrt{5 x})}^{2} = {(- 5)}^{2}$ $(sqrt(5x))^2 = (-5)^2$

$5 x = 25$ $5x = 25$

$x = 5$ $x = 5$

$5$ $5$ is not a solution to the original equation, but is is a solution to the equation: $5 x = 25$ $5x = 25$ .

This is one way that extraneous solutions can be introduced in the process of solving an equation.

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How do you solve and check for extraneous solutions in $\sqrt{5 x} = - 5$ $sqrt(5x) = -5$ ?

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How do you solve and check for extraneous solutions in √5x=−5sqrt(5x) = -5?

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How do you solve and check for extraneous solutions in $\sqrt{5 x} = - 5$ $sqrt(5x) = -5$ ?