Solving Radical Equations

$x = 14$ does not solve the original equation. We could have expected this from the beginning because the square root of a number can never be negative. Because of this, the equation lacks real solutions.

c

We proceed in the same manner with the last equation.

60 - 3 x = 9

RaiseEqn

LHS^{2} = RHS^{2}

60 - 3 x = 9^{2}

CalcPowCalculate power

60 - 3 x = 81

SubEqn

LHS - 60 = RHS - 60

- 3 x = 21

DivEqn

LHS / (- 3) = RHS / (- 3)

x = - 7

Next we need to test if

x = - 7

is a solution.

60 - 3 x = 9

Substitute

x = - 7

60 - 3 (- 7) = ? 9

MultNegNegOnePar

- a (- b) = a \cdot b

60 + 21 = ? 9

AddTermsAdd terms

81 = ? 9

CalcRootCalculate root

9 = 9

Since we arrived at a true statement, we know that

x = - 7

is a solution to the equation.

Solving Radical Equations 1.2 - Solution