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Solving Radical Equations

Solving Radical Equations 1.20 - Solution

arrow_back Return to Solving Radical Equations
To solve equations with a variable expression inside a radical, we first want to make sure the radical is isolated. Then we can raise both sides of the equation to a power equal to the index of the radical. Let's try to solve our equation using this method!
x2+3=x+1\sqrt{x^2+3}=x+1
(x2+3)2=(x+1)2\left(\sqrt{x^2+3}\right)^2=(x+1)^2
x2+3=(x+1)2x^2+3 = (x+1)^2
x2+3=x2+2x(1)+12x^2+3 = x^2+2x(1)+1^2
x2+3=x2+2x+1x^2+3 = x^2+2x+1
Solve for xx
3=2x+13 = 2x+1
2=2x2 = 2x
1=x1=x
x=1x=1
Next, we will check for extraneous solutions. We do that by substituting 11 for xx 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.
x2+3=x+1\sqrt{x^2+3}=x+1
12+3=?1+1\sqrt{{\color{#0000FF}{1}}^2+3}\stackrel{?}{=}{\color{#0000FF}{1}}+1
1+3=?1+1\sqrt{1+3}\stackrel{?}{=}1+1
4=?2\sqrt{4}\stackrel{?}{=}2
2=2 2=2 \ {\color{#009600}{\huge{\checkmark}}}
Because our substitution produced a true statement, we know that our answer, x=1,x=1, is correct.