Radical equations are equations that include square roots, cube roots, or rational exponents. To solve them, isolate the radical when possible, raise both sides to the correct power, and then check for extraneous solutions. These problems include one-radical equations, two-radical equations, cube root equations, rational exponent equations, and equations with no solution.
Problems & Videos
Solve for x.
\(\textbf{1)}\) \( \sqrt{3x+1}-4=0 \)
The answer is \( x=5 \)
\(\,\,\,\,\,\sqrt{3x+1}-4=0\)
\(\,\,\,\,\,\sqrt{3x+1}=4\)
\(\,\,\,\,\,\left(\sqrt{3x+1}\right)^2=4^2\)
\(\,\,\,\,\,3x+1=16\)
\(\,\,\,\,\,3x=15\)
\(\,\,\,\,\,x=5\)
\(\,\,\,\,\,\)The answer is \(x=5\)
\(\textbf{2)}\) \( 3\sqrt{x+1}=12 \)
The answer is \( x=15 \)
\(\,\,\,\,\,3\sqrt{x+1}=12\)
\(\,\,\,\,\,\sqrt{x+1}=4\)
\(\,\,\,\,\,\left(\sqrt{x+1}\right)^2=4^2\)
\(\,\,\,\,\,x+1=16\)
\(\,\,\,\,\,x=15\)
\(\,\,\,\,\,\)The answer is \(x=15\)
\(\textbf{3)}\) \( \sqrt{x+1}+\sqrt{x-6}=7 \)
The answer is \( x=15 \)
\(\,\,\,\,\,\sqrt{x+1}+\sqrt{x-6}=7\)
\(\,\,\,\,\,\sqrt{x+1}=7-\sqrt{x-6}\)
\(\,\,\,\,\,\left(\sqrt{x+1}\right)^2=\left(7-\sqrt{x-6}\right)^2\)
\(\,\,\,\,\,x+1=49-14\sqrt{x-6}+x-6\)
\(\,\,\,\,\,x+1=x+43-14\sqrt{x-6}\)
\(\,\,\,\,\,-42=-14\sqrt{x-6}\)
\(\,\,\,\,\,3=\sqrt{x-6}\)
\(\,\,\,\,\,9=x-6\)
\(\,\,\,\,\,x=15\)
\(\,\,\,\,\,\)The answer is \(x=15\)
\(\textbf{4)}\) \( \sqrt{x+1}=\sqrt{2x-7} \)
The answer is \( x=8 \)
\(\,\,\,\,\,\sqrt{x+1}=\sqrt{2x-7}\)
\(\,\,\,\,\,\left(\sqrt{x+1}\right)^2=\left(\sqrt{2x-7}\right)^2\)
\(\,\,\,\,\,x+1=2x-7\)
\(\,\,\,\,\,8=x\)
\(\,\,\,\,\,x=8\)
\(\,\,\,\,\,\)The answer is \(x=8\)
\(\textbf{5)}\) \( \sqrt[3]{x+1}=2 \)
The answer is \( x=7 \)
\(\,\,\,\,\,\sqrt[3]{x+1}=2\)
\(\,\,\,\,\,\left(\sqrt[3]{x+1}\right)^3=2^3\)
\(\,\,\,\,\,x+1=8\)
\(\,\,\,\,\,x=7\)
\(\,\,\,\,\,\)The answer is \(x=7\)
\(\textbf{6)}\) \( (x-5)^{(3/2)}=27 \)
The answer is \( x=14 \)
\(\,\,\,\,\,(x-5)^{(3/2)}=27\)
\(\,\,\,\,\,\left(\sqrt{x-5}\right)^3=27\)
\(\,\,\,\,\,\sqrt{x-5}=3\)
\(\,\,\,\,\,\left(\sqrt{x-5}\right)^2=3^2\)
\(\,\,\,\,\,x-5=9\)
\(\,\,\,\,\,x=14\)
\(\,\,\,\,\,\)The answer is \(x=14\)
\(\textbf{7)}\) \( \sqrt{x+1}=-5 \)
The answer is No solution
\(\,\,\,\,\,\sqrt{x+1}=-5\)
\(\,\,\,\,\,\sqrt{x+1}\) is never negative for real numbers.
\(\,\,\,\,\,\sqrt{x+1}=-5\) cannot be true.
\(\,\,\,\,\,\)The answer is No solution.
\(\textbf{8)}\) \( \sqrt{x+5}=\sqrt{x}+1 \)
The answer is \( x=4\)
\(\,\,\,\,\,\sqrt{x+5}=\sqrt{x}+1\)
\(\,\,\,\,\,\left(\sqrt{x+5}\right)^2=\left(\sqrt{x}+1\right)^2\)
\(\,\,\,\,\,x+5=x+2\sqrt{x}+1\)
\(\,\,\,\,\,4=2\sqrt{x}\)
\(\,\,\,\,\,2=\sqrt{x}\)
\(\,\,\,\,\,2^2=\left(\sqrt{x}\right)^2\)
\(\,\,\,\,\,4=x\)
\(\,\,\,\,\,\)The answer is \(x=4\)
\(\textbf{9)}\) \(10-\sqrt{y}=\frac{4}{5}\sqrt{y}+1\)
The answer is \( y=25\)
\(\,\,\,\,\,10-\sqrt{y}=\frac{4}{5}\sqrt{y}+1\)
\(\,\,\,\,\,9-\sqrt{y}=\frac{4}{5}\sqrt{y}\)
\(\,\,\,\,\,9=\frac{9}{5}\sqrt{y}\)
\(\,\,\,\,\,5=\sqrt{y}\)
\(\,\,\,\,\,5^2=\left(\sqrt{y}\right)^2\)
\(\,\,\,\,\,25=y\)
\(\,\,\,\,\,\)The answer is \(y=25\)
\(\textbf{10)}\) \(\sqrt{x-5}=5-\sqrt{x}\)
The answer is \( x=9\)
\(\,\,\,\,\,\sqrt{x-5}=5-\sqrt{x}\)
\(\,\,\,\,\,\left(\sqrt{x-5}\right)^2=\left(5-\sqrt{x}\right)^2\)
\(\,\,\,\,\,x-5=25-10\sqrt{x}+x\)
\(\,\,\,\,\,-30=-10\sqrt{x}\)
\(\,\,\,\,\,3=\sqrt{x}\)
\(\,\,\,\,\,9=x\)
\(\,\,\,\,\,\)The answer is \(x=9\)
\(\textbf{11)}\) \(\sqrt{x-1}=1-\sqrt{x}\)
The answer is \( x=1\)
\(\,\,\,\,\,\sqrt{x-1}=1-\sqrt{x}\)
\(\,\,\,\,\,\left(\sqrt{x-1}\right)^2=\left(1-\sqrt{x}\right)^2\)
\(\,\,\,\,\,x-1=1-2\sqrt{x}+x\)
\(\,\,\,\,\,-2=-2\sqrt{x}\)
\(\,\,\,\,\,1=\sqrt{x}\)
\(\,\,\,\,\,1=x\)
\(\,\,\,\,\,\)The answer is \(x=1\)
\(\textbf{12)}\) \(x=\sqrt{x+2}\)
The answer is \( x=2\)
\(\,\,\,\,\,x=\sqrt{x+2}\)
\(\,\,\,\,\,x^2=x+2\)
\(\,\,\,\,\,x^2-x-2=0\)
\(\,\,\,\,\,(x-2)(x+1)=0\)
\(\,\,\,\,\,x=2\text{ or }x=-1\)
\(\,\,\,\,\,x=-1\) does not check in the original equation.
\(\,\,\,\,\,\)The answer is \(x=2\)
\(\textbf{13)}\) \(\sqrt{2x-3}=5\)
The answer is \(x=14\)
\(\,\,\,\,\,\sqrt{2x-3}=5\)
\(\,\,\,\,\,\left(\sqrt{2x-3}\right)^2=5^2\)
\(\,\,\,\,\,2x-3=25\)
\(\,\,\,\,\,2x=28\)
\(\,\,\,\,\,x=14\)
\(\,\,\,\,\,\)The answer is \(x=14\)
\(\textbf{14)}\) \(\sqrt{x-4}+2=7\)
The answer is \(x=29\)
\(\,\,\,\,\,\sqrt{x-4}+2=7\)
\(\,\,\,\,\,\sqrt{x-4}=5\)
\(\,\,\,\,\,\left(\sqrt{x-4}\right)^2=5^2\)
\(\,\,\,\,\,x-4=25\)
\(\,\,\,\,\,x=29\)
\(\,\,\,\,\,\)The answer is \(x=29\)
\(\textbf{15)}\) \(2\sqrt{x-1}=8\)
The answer is \(x=17\)
\(\,\,\,\,\,2\sqrt{x-1}=8\)
\(\,\,\,\,\,\sqrt{x-1}=4\)
\(\,\,\,\,\,\left(\sqrt{x-1}\right)^2=4^2\)
\(\,\,\,\,\,x-1=16\)
\(\,\,\,\,\,x=17\)
\(\,\,\,\,\,\)The answer is \(x=17\)
\(\textbf{16)}\) \(\sqrt[3]{2x-1}=3\)
The answer is \(x=14\)
\(\,\,\,\,\,\sqrt[3]{2x-1}=3\)
\(\,\,\,\,\,\left(\sqrt[3]{2x-1}\right)^3=3^3\)
\(\,\,\,\,\,2x-1=27\)
\(\,\,\,\,\,2x=28\)
\(\,\,\,\,\,x=14\)
\(\,\,\,\,\,\)The answer is \(x=14\)
\(\textbf{17)}\) \((x+2)^{(1/2)}=6\)
The answer is \(x=34\)
\(\,\,\,\,\,(x+2)^{(1/2)}=6\)
\(\,\,\,\,\,\sqrt{x+2}=6\)
\(\,\,\,\,\,\left(\sqrt{x+2}\right)^2=6^2\)
\(\,\,\,\,\,x+2=36\)
\(\,\,\,\,\,x=34\)
\(\,\,\,\,\,\)The answer is \(x=34\)
\(\textbf{18)}\) \(\sqrt{x+9}=x-3\)
The answer is \(x=7\)
\(\,\,\,\,\,\sqrt{x+9}=x-3\)
\(\,\,\,\,\,\left(\sqrt{x+9}\right)^2=(x-3)^2\)
\(\,\,\,\,\,x+9=x^2-6x+9\)
\(\,\,\,\,\,0=x^2-7x\)
\(\,\,\,\,\,0=x(x-7)\)
\(\,\,\,\,\,x=0\text{ or }x=7\)
\(\,\,\,\,\,x=0\) does not check in the original equation.
\(\,\,\,\,\,\)The answer is \(x=7\)
\(\textbf{19)}\) \(\sqrt{x+4}=x-2\)
The answer is \(x=5\)
\(\,\,\,\,\,\sqrt{x+4}=x-2\)
\(\,\,\,\,\,\left(\sqrt{x+4}\right)^2=(x-2)^2\)
\(\,\,\,\,\,x+4=x^2-4x+4\)
\(\,\,\,\,\,0=x^2-5x\)
\(\,\,\,\,\,0=x(x-5)\)
\(\,\,\,\,\,x=0\text{ or }x=5\)
\(\,\,\,\,\,x=0\) does not check in the original equation.
\(\,\,\,\,\,\)The answer is \(x=5\)
\(\textbf{20)}\) \(\sqrt{x+6}-\sqrt{x-3}=1\)
The answer is \(x=7\)
\(\,\,\,\,\,\sqrt{x+6}-\sqrt{x-3}=1\)
\(\,\,\,\,\,\sqrt{x+6}=1+\sqrt{x-3}\)
\(\,\,\,\,\,\left(\sqrt{x+6}\right)^2=\left(1+\sqrt{x-3}\right)^2\)
\(\,\,\,\,\,x+6=1+2\sqrt{x-3}+x-3\)
\(\,\,\,\,\,x+6=x-2+2\sqrt{x-3}\)
\(\,\,\,\,\,8=2\sqrt{x-3}\)
\(\,\,\,\,\,4=\sqrt{x-3}\)
\(\,\,\,\,\,16=x-3\)
\(\,\,\,\,\,x=19\)
\(\,\,\,\,\,\)The answer is \(x=19\)
Notes
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