Why does solving radical equations need a different process? how do extraneous solutions arise from radical equations?

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.

i.e.  ( Adding 3 on both sides )

i.e.

i.e.

i.e. .

So, the solution the the radical equation  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

i.e.

i.e.

i.e.

i.e.

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

Substituting x = 0 in , gives  i.e.  i.e. , which is not possible.

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

Hence, x = 0 is an extraneous solution.

Step-by-step explanation:

answer

d. 10

explanation

the distance from point y to wx is the height of the triangle.

using pythagoras theorem,

group similar terms to get,

this implies that,

take positive square root of both sides to get,

this implies that,

1. perpendicular

2. parallel

3. perpendicular