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Raise each side of the equation to a power equal to the index of the radical to eliminate the radical.
x=3
Solving a radical equation usually involves three main steps.
Now we can analyze the given radical equation.
sqrt(2x+30)=x+3
LHS^2=RHS^2
( sqrt(a) )^2 = a
(a+b)^2=a^2+2ab+b^2
Multiply
Calculate power
LHS-30=RHS-30
LHS+2x=RHS+2x
Rearrange equation
We obtained a quadratic equation. There are many ways to solve a quadratic equation. We will use the Quadratic Formula to solve this equation. Let's determine a, b, and c. x^2+4x-21=0 ⇔ 1x^2+ 4x+( - 21)=0 We see above that a= 1, b= 4, and c= - 21. Let's substitute these values into the Quadratic Formula to solve the equation.
Substitute values
Now we can calculate the first root using the positive sign and the second root using the negative sign.
| x=- 4± 10/2 | |
|---|---|
| x=- 4+10/2 | x=- 4-10/2 |
| x=3 | x=- 7 |
Next, we will check the solutions by substituting 3 and - 7 for x into the original equation. If the substitution produces a true statement, we know that our answer is correct. If it does not, then it is an extraneous solution. Let's first check the 3.
Because our substitution produced a true statement, we know that our answer, x=3, is correct. Let's check now the - 7.
Because our substitution produced a false statement, we know that our answer, x=- 7, is not correct.