Question

What reasoning and explanations can be used when solving radical equations and how do extraneous solutions arise from radical equations?

Answer 1

Answer:

We know that, a 'radical equation' is an equation that contains radical expressions, which further are the expressions containing radicals ( square roots and other roots of numbers ).

In order to solve radical equations, we use the rules of exponents and basic algebraic properties.

The common reasoning to use while solving a radical equation is:

1. Isolate the radical expression.

2. Square both sides of the equation to remove radical.

3. After removing the radical, solve the equation to find the unkown variable

4. Check the answer for the errors occurred by removing the radicals.

For e.g. $\sqrt{x}-3=5$

i.e. $\sqrt{x}-3+3=5+3$ ( Adding 3 on both sides )

i.e. $\sqrt{x}=8$

i.e. $(\sqrt{x})^{2} =8^{2}$

i.e. $x=64$.

So, the solution the the radical equation $\sqrt{x}-3=5$ is x = 64.

Further, we know that an 'extraneous solution' is that root of the radical equation which is not a root of the original equation and is excluded from the domain.

for e.g. Take $\sqrt{x+4} =x-2$

i.e. $(\sqrt{x+4})^{2} =(x-2)^2$

i.e. $x+4=x^2+4-4x$

i.e. $0=x^2-5x$

i.e. $0=x(x-5)$

i.e. x = 0 and x = 5.

Substituting x = 0 in $\sqrt{x+4} =x-2$, gives $\sqrt{0+4} =0-2$ i.e. $\sqrt{4} =-2$ i.e. $2=-2$, which is not possible.

So, x = 0 is a solution that does not satisfy the equation.

Hence, x = 0 is an extraneous solution.