Equations involving squared variables can often be solved using the square root method. First, isolate the squared expression, then take the square root of both sides, remembering to include both positive and negative solutions when appropriate. This method works for equations involving perfect squares, fractions, radicals, and squared expressions containing parentheses.
Practice Problems
\(\textbf{1)}\) \(x^2=36\) The answer is \(x=\pm6\)
\(\,\,\,\,\,\, x^2=36\)
\(\,\,\,\,\,\, x=\pm\sqrt{36}\)
\(\,\,\,\,\,\, x=\pm6\)
\(\,\,\,\,\,\, x^2=36\)
\(\,\,\,\,\,\, x=\pm\sqrt{36}\)
\(\,\,\,\,\,\, x=\pm6\)
\(\textbf{2)}\) \(x^2+5=14\) The answer is \(x=\pm3\)
\(\,\,\,\,\,\, x^2+5=14\)
\(\,\,\,\,\,\, x^2=9\)
\(\,\,\,\,\,\, x=\pm\sqrt{9}\)
\(\,\,\,\,\,\, x=\pm3\)
\(\,\,\,\,\,\, x^2+5=14\)
\(\,\,\,\,\,\, x^2=9\)
\(\,\,\,\,\,\, x=\pm\sqrt{9}\)
\(\,\,\,\,\,\, x=\pm3\)
\(\textbf{3)}\) \(2x^2-6=26\) The answer is \(x=\pm4\)
\(\,\,\,\,\,\, 2x^2-6=26\)
\(\,\,\,\,\,\, 2x^2=32\)
\(\,\,\,\,\,\, x^2=16\)
\(\,\,\,\,\,\, x=\pm\sqrt{16}\)
\(\,\,\,\,\,\, x=\pm4\)
\(\,\,\,\,\,\, 2x^2-6=26\)
\(\,\,\,\,\,\, 2x^2=32\)
\(\,\,\,\,\,\, x^2=16\)
\(\,\,\,\,\,\, x=\pm\sqrt{16}\)
\(\,\,\,\,\,\, x=\pm4\)
\(\textbf{4)}\) \(\left(x-5\right)^2=25\) The answer is \(x=0\text{ or }x=10\)
\(\,\,\,\,\,\, \left(x-5\right)^2=25\)
\(\,\,\,\,\,\, x-5=\pm\sqrt{25}\)
\(\,\,\,\,\,\, x-5=\pm5\)
\(\,\,\,\,\,\, x=5\pm5\)
\(\,\,\,\,\,\, x=5-5\text{ or }x=5+5\)
\(\,\,\,\,\,\, x=0\text{ or }x=10\)
\(\,\,\,\,\,\, \left(x-5\right)^2=25\)
\(\,\,\,\,\,\, x-5=\pm\sqrt{25}\)
\(\,\,\,\,\,\, x-5=\pm5\)
\(\,\,\,\,\,\, x=5\pm5\)
\(\,\,\,\,\,\, x=5-5\text{ or }x=5+5\)
\(\,\,\,\,\,\, x=0\text{ or }x=10\)
\(\textbf{5)}\) \(3\left(x-2\right)^2+1=4\) The answer is \(x=1\text{ or }x=3\)
\(\,\,\,\,\,\, 3\left(x-2\right)^2+1=4\)
\(\,\,\,\,\,\, 3\left(x-2\right)^2=3\)
\(\,\,\,\,\,\, \left(x-2\right)^2=1\)
\(\,\,\,\,\,\, x-2=\pm\sqrt{1}\)
\(\,\,\,\,\,\, x-2=\pm1\)
\(\,\,\,\,\,\, x=2\pm1\)
\(\,\,\,\,\,\, x=1\text{ or }x=3\)
\(\,\,\,\,\,\, 3\left(x-2\right)^2+1=4\)
\(\,\,\,\,\,\, 3\left(x-2\right)^2=3\)
\(\,\,\,\,\,\, \left(x-2\right)^2=1\)
\(\,\,\,\,\,\, x-2=\pm\sqrt{1}\)
\(\,\,\,\,\,\, x-2=\pm1\)
\(\,\,\,\,\,\, x=2\pm1\)
\(\,\,\,\,\,\, x=1\text{ or }x=3\)
\(\textbf{6)}\) \(2\left(x+1\right)^2+3=11\) The answer is \(x=1\text{ or }x=-3\)
\(\,\,\,\,\,\, 2\left(x+1\right)^2+3=11\)
\(\,\,\,\,\,\, 2\left(x+1\right)^2=8\)
\(\,\,\,\,\,\, \left(x+1\right)^2=4\)
\(\,\,\,\,\,\, x+1=\pm\sqrt{4}\)
\(\,\,\,\,\,\, x+1=\pm2\)
\(\,\,\,\,\,\, x=-1\pm2\)
\(\,\,\,\,\,\, x=-3\text{ or }x=1\)
\(\,\,\,\,\,\, 2\left(x+1\right)^2+3=11\)
\(\,\,\,\,\,\, 2\left(x+1\right)^2=8\)
\(\,\,\,\,\,\, \left(x+1\right)^2=4\)
\(\,\,\,\,\,\, x+1=\pm\sqrt{4}\)
\(\,\,\,\,\,\, x+1=\pm2\)
\(\,\,\,\,\,\, x=-1\pm2\)
\(\,\,\,\,\,\, x=-3\text{ or }x=1\)
\(\textbf{7)}\) \(x^2=81\) The answer is \(x=\pm9\)
\(\,\,\,\,\,\, x^2=81\)
\(\,\,\,\,\,\, x=\pm\sqrt{81}\)
\(\,\,\,\,\,\, x=\pm9\)
\(\,\,\,\,\,\, x^2=81\)
\(\,\,\,\,\,\, x=\pm\sqrt{81}\)
\(\,\,\,\,\,\, x=\pm9\)
\(\textbf{8)}\) \(x^2-12=13\) The answer is \(x=\pm5\)
\(\,\,\,\,\,\, x^2-12=13\)
\(\,\,\,\,\,\, x^2=25\)
\(\,\,\,\,\,\, x=\pm\sqrt{25}\)
\(\,\,\,\,\,\, x=\pm5\)
\(\,\,\,\,\,\, x^2-12=13\)
\(\,\,\,\,\,\, x^2=25\)
\(\,\,\,\,\,\, x=\pm\sqrt{25}\)
\(\,\,\,\,\,\, x=\pm5\)
\(\textbf{9)}\) \(5x^2+7=87\) The answer is \(x=\pm4\)
\(\,\,\,\,\,\, 5x^2+7=87\)
\(\,\,\,\,\,\, 5x^2=80\)
\(\,\,\,\,\,\, x^2=16\)
\(\,\,\,\,\,\, x=\pm\sqrt{16}\)
\(\,\,\,\,\,\, x=\pm4\)
\(\,\,\,\,\,\, 5x^2+7=87\)
\(\,\,\,\,\,\, 5x^2=80\)
\(\,\,\,\,\,\, x^2=16\)
\(\,\,\,\,\,\, x=\pm\sqrt{16}\)
\(\,\,\,\,\,\, x=\pm4\)
\(\textbf{10)}\) \(\left(x+4\right)^2=49\) The answer is \(x=3\text{ or }x=-11\)
\(\,\,\,\,\,\, \left(x+4\right)^2=49\)
\(\,\,\,\,\,\, x+4=\pm\sqrt{49}\)
\(\,\,\,\,\,\, x+4=\pm7\)
\(\,\,\,\,\,\, x=-4\pm7\)
\(\,\,\,\,\,\, x=3\text{ or }x=-11\)
\(\,\,\,\,\,\, \left(x+4\right)^2=49\)
\(\,\,\,\,\,\, x+4=\pm\sqrt{49}\)
\(\,\,\,\,\,\, x+4=\pm7\)
\(\,\,\,\,\,\, x=-4\pm7\)
\(\,\,\,\,\,\, x=3\text{ or }x=-11\)
\(\textbf{11)}\) \(4\left(x-3\right)^2=100\) The answer is \(x=-2\text{ or }x=8\)
\(\,\,\,\,\,\, 4\left(x-3\right)^2=100\)
\(\,\,\,\,\,\, \left(x-3\right)^2=25\)
\(\,\,\,\,\,\, x-3=\pm\sqrt{25}\)
\(\,\,\,\,\,\, x-3=\pm5\)
\(\,\,\,\,\,\, x=3\pm5\)
\(\,\,\,\,\,\, x=-2\text{ or }x=8\)
\(\,\,\,\,\,\, 4\left(x-3\right)^2=100\)
\(\,\,\,\,\,\, \left(x-3\right)^2=25\)
\(\,\,\,\,\,\, x-3=\pm\sqrt{25}\)
\(\,\,\,\,\,\, x-3=\pm5\)
\(\,\,\,\,\,\, x=3\pm5\)
\(\,\,\,\,\,\, x=-2\text{ or }x=8\)
\(\textbf{12)}\) \(\left(2x-1\right)^2=49\) The answer is \(x=4\text{ or }x=-3\)
\(\,\,\,\,\,\, \left(2x-1\right)^2=49\)
\(\,\,\,\,\,\, 2x-1=\pm\sqrt{49}\)
\(\,\,\,\,\,\, 2x-1=\pm7\)
\(\,\,\,\,\,\, 2x=1\pm7\)
\(\,\,\,\,\,\, x=\frac{1\pm7}{2}\)
\(\,\,\,\,\,\, x=4\text{ or }x=-3\)
\(\,\,\,\,\,\, \left(2x-1\right)^2=49\)
\(\,\,\,\,\,\, 2x-1=\pm\sqrt{49}\)
\(\,\,\,\,\,\, 2x-1=\pm7\)
\(\,\,\,\,\,\, 2x=1\pm7\)
\(\,\,\,\,\,\, x=\frac{1\pm7}{2}\)
\(\,\,\,\,\,\, x=4\text{ or }x=-3\)
\(\textbf{13)}\) \(2\left(3x+1\right)^2-5=13\) The answer is \(x=\frac{2}{3}\text{ or }x=-\frac{4}{3}\)
\(\,\,\,\,\,\, 2\left(3x+1\right)^2-5=13\)
\(\,\,\,\,\,\, 2\left(3x+1\right)^2=18\)
\(\,\,\,\,\,\, \left(3x+1\right)^2=9\)
\(\,\,\,\,\,\, 3x+1=\pm\sqrt{9}\)
\(\,\,\,\,\,\, 3x+1=\pm3\)
\(\,\,\,\,\,\, 3x=-1\pm3\)
\(\,\,\,\,\,\, x=\frac{-1\pm3}{3}\)
\(\,\,\,\,\,\, x=\frac{2}{3}\text{ or }x=-\frac{4}{3}\)
\(\,\,\,\,\,\, 2\left(3x+1\right)^2-5=13\)
\(\,\,\,\,\,\, 2\left(3x+1\right)^2=18\)
\(\,\,\,\,\,\, \left(3x+1\right)^2=9\)
\(\,\,\,\,\,\, 3x+1=\pm\sqrt{9}\)
\(\,\,\,\,\,\, 3x+1=\pm3\)
\(\,\,\,\,\,\, 3x=-1\pm3\)
\(\,\,\,\,\,\, x=\frac{-1\pm3}{3}\)
\(\,\,\,\,\,\, x=\frac{2}{3}\text{ or }x=-\frac{4}{3}\)
\(\textbf{14)}\) \(x^2=20\) The answer is \(x=\pm2\sqrt{5}\)
\(\,\,\,\,\,\, x^2=20\)
\(\,\,\,\,\,\, x=\pm\sqrt{20}\)
\(\,\,\,\,\,\, x=\pm\sqrt{4\cdot5}\)
\(\,\,\,\,\,\, x=\pm2\sqrt{5}\)
\(\,\,\,\,\,\, x^2=20\)
\(\,\,\,\,\,\, x=\pm\sqrt{20}\)
\(\,\,\,\,\,\, x=\pm\sqrt{4\cdot5}\)
\(\,\,\,\,\,\, x=\pm2\sqrt{5}\)
\(\textbf{15)}\) \(3x^2-4=20\) The answer is \(x=\pm2\sqrt{2}\)
\(\,\,\,\,\,\, 3x^2-4=20\)
\(\,\,\,\,\,\, 3x^2=24\)
\(\,\,\,\,\,\, x^2=8\)
\(\,\,\,\,\,\, x=\pm\sqrt{8}\)
\(\,\,\,\,\,\, x=\pm\sqrt{4\cdot2}\)
\(\,\,\,\,\,\, x=\pm2\sqrt{2}\)
\(\,\,\,\,\,\, 3x^2-4=20\)
\(\,\,\,\,\,\, 3x^2=24\)
\(\,\,\,\,\,\, x^2=8\)
\(\,\,\,\,\,\, x=\pm\sqrt{8}\)
\(\,\,\,\,\,\, x=\pm\sqrt{4\cdot2}\)
\(\,\,\,\,\,\, x=\pm2\sqrt{2}\)
\(\textbf{16)}\) \(\left(x+2\right)^2=12\) The answer is \(x=-2\pm2\sqrt{3}\)
\(\,\,\,\,\,\, \left(x+2\right)^2=12\)
\(\,\,\,\,\,\, x+2=\pm\sqrt{12}\)
\(\,\,\,\,\,\, x+2=\pm\sqrt{4\cdot3}\)
\(\,\,\,\,\,\, x+2=\pm2\sqrt{3}\)
\(\,\,\,\,\,\, x=-2\pm2\sqrt{3}\)
\(\,\,\,\,\,\, \left(x+2\right)^2=12\)
\(\,\,\,\,\,\, x+2=\pm\sqrt{12}\)
\(\,\,\,\,\,\, x+2=\pm\sqrt{4\cdot3}\)
\(\,\,\,\,\,\, x+2=\pm2\sqrt{3}\)
\(\,\,\,\,\,\, x=-2\pm2\sqrt{3}\)
\(\textbf{17)}\) \(x^2=\frac{9}{16}\) The answer is \(x=\pm\frac{3}{4}\)
\(\,\,\,\,\,\, x^2=\frac{9}{16}\)
\(\,\,\,\,\,\, x=\pm\sqrt{\frac{9}{16}}\)
\(\,\,\,\,\,\, x=\pm\frac{\sqrt{9}}{\sqrt{16}}\)
\(\,\,\,\,\,\, x=\pm\frac{3}{4}\)
\(\,\,\,\,\,\, x^2=\frac{9}{16}\)
\(\,\,\,\,\,\, x=\pm\sqrt{\frac{9}{16}}\)
\(\,\,\,\,\,\, x=\pm\frac{\sqrt{9}}{\sqrt{16}}\)
\(\,\,\,\,\,\, x=\pm\frac{3}{4}\)
\(\textbf{18)}\) \(\left(x-\frac{1}{2}\right)^2=\frac{25}{4}\) The answer is \(x=3\text{ or }x=-2\)
\(\,\,\,\,\,\, \left(x-\frac{1}{2}\right)^2=\frac{25}{4}\)
\(\,\,\,\,\,\, x-\frac{1}{2}=\pm\sqrt{\frac{25}{4}}\)
\(\,\,\,\,\,\, x-\frac{1}{2}=\pm\frac{5}{2}\)
\(\,\,\,\,\,\, x=\frac{1}{2}\pm\frac{5}{2}\)
\(\,\,\,\,\,\, x=\frac{1+5}{2}\text{ or }x=\frac{1-5}{2}\)
\(\,\,\,\,\,\, x=3\text{ or }x=-2\)
\(\,\,\,\,\,\, \left(x-\frac{1}{2}\right)^2=\frac{25}{4}\)
\(\,\,\,\,\,\, x-\frac{1}{2}=\pm\sqrt{\frac{25}{4}}\)
\(\,\,\,\,\,\, x-\frac{1}{2}=\pm\frac{5}{2}\)
\(\,\,\,\,\,\, x=\frac{1}{2}\pm\frac{5}{2}\)
\(\,\,\,\,\,\, x=\frac{1+5}{2}\text{ or }x=\frac{1-5}{2}\)
\(\,\,\,\,\,\, x=3\text{ or }x=-2\)
\(\textbf{19)}\) \(\left(x-3\right)^2=0\) The answer is \(x=3\)
\(\,\,\,\,\,\, \left(x-3\right)^2=0\)
\(\,\,\,\,\,\, x-3=\sqrt{0}\)
\(\,\,\,\,\,\, x-3=0\)
\(\,\,\,\,\,\, x=3\)
\(\,\,\,\,\,\, \left(x-3\right)^2=0\)
\(\,\,\,\,\,\, x-3=\sqrt{0}\)
\(\,\,\,\,\,\, x-3=0\)
\(\,\,\,\,\,\, x=3\)
\(\textbf{20)}\) \(2\left(x+4\right)^2+5=3\) There is no real solution.
\(\,\,\,\,\,\, 2\left(x+4\right)^2+5=3\)
\(\,\,\,\,\,\, 2\left(x+4\right)^2=-2\)
\(\,\,\,\,\,\, \left(x+4\right)^2=-1\)
A real number squared cannot be negative.
Therefore, there is no real solution.
\(\,\,\,\,\,\, 2\left(x+4\right)^2+5=3\)
\(\,\,\,\,\,\, 2\left(x+4\right)^2=-2\)
\(\,\,\,\,\,\, \left(x+4\right)^2=-1\)
A real number squared cannot be negative.
Therefore, there is no real solution.
See Related Pages\(\)
\(\bullet\text{ Solving Equations by Addition and Subtraction}\)
\(\,\,\,\,\,\,\,\,x+3=4…\)
\(\bullet\text{ Solving Equations by Multiplication and Division}\)
\(\,\,\,\,\,\,\,\,8x=48…\)
\(\bullet\text{ Solving Multi-step Equations}\)
\(\,\,\,\,\,\,\,\,3x+2=14…\)
\(\bullet\text{ Solving Equations with Variables on Both Sides}\)
\(\,\,\,\,\,\,\,\,3x+5=7x-3…\)
\(\bullet\text{ Solving Equations with Decimals}\)
\(\,\,\,\,\,\,\,\,43.5+0.2x=51.1…\)
\(\bullet\text{ Solving Equations with Fractions}\)
\(\,\,\,\,\,\,\,\,\frac{2}{5}x+\frac{2}{3}=\frac{8}{3}…\)