R adical equation A radical equation is an equation in which the variable has fractional form or fractional exponent and occurs in a radicand.

How to solve the radical equation:

How to solve the radical equation The radical must be eliminated or the fractional form must be made an integral exponent. C hecking the obtained solution with the original equation is very important.

Example 1::

Example 1: ( = 4^ 2 Square both sides of the equation to eliminate the radical equation. 5x-9 = 16 Transpose 9 to eliminate negative sign 5x = 16 + 9 Then add . 5x = 25 X = 5 Find the square of 25 Your x is 5.

Checking::

Checking: = 4 = 4 = 4 4 = 4 True , your x is 4. Find the square root of 16.

Example 2: :

Example 2: ( √2x-3) 3 = ( 3 √ x+1 ) 3 2x – 3 = x + 1 2x – x = 1 + 3 x = 4 Raise both sides of the equation to cube, to cancel the radical sign. Combine like terms Your x is 4.

Checking::

Checking: 3 √2(4) – 3 = 3 √8 – 3 3 √8 – 3 = 3 √5 3 √5 = 3 √5 Multiply 4 x 2 . And subtract 8 – 3. Same number , same radical sign. Your answer is correct .

Example 3::

Example 3: Solve: x + 2 = √2x+28 Solution: ( x+2) 2 = (√2x+28) 2 X 2 + 4x + 4 = 2x + 28 X 2 + 4x – 2x + 4 -28 = 0 X 2 + 2x – 24 = 0 (x + 6)(x + 4) = 0 x + 6 = 0 X – 4 = 0 Square both sides of the equation. Combine like terms , variable to variable. x = -6 transpose 6 to eliminate the positive sign x = 4 transpose to eliminate the negative sign .

Checking::

Checking: For , x = -6 -6 + 2 = √2(-6) + 28 -4 = √-12 + 28 -4 = 16 -4 = 4 -6 is not solution to the equation. Now , it is called extraneous root .

Checking: :

Checking: For x = 4 Now , let’s try this! 4 + 2 = √2(4) + 28 6 = √8 + 28 6 = √36 6 = 6 Find the square root of 36. True , x = 4 is only the solution to the equation .

Teacher::

Teacher: Ms. Joana Marie Padilla

Grade and section::

Grade and section: IX – Eduardo San Juan

Group leader::

Group leader: Krystle Callao Macaspac group 6

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