Practice Problems
\(\textbf{1)}\) \(x^2=36\)
The answer is \( x=\pm6 \)
\(\,\,\,\,\,\, x^2=36 \)
\(\,\,\,\,\,\, \sqrt{x^2}=\pm\sqrt{36}\)
\(\,\,\,\,\,\, x=\pm 6\)
\(\textbf{2)}\) \(x^2+5=14\)
The answer is \( x=\pm3 \)
\(\,\,\,\,\,\, x^2+5=14 \)
\(\,\,\,\,\,\, x^2=9\)
\(\,\,\,\,\,\, \sqrt{x^2}=\pm\sqrt{9}\)
\(\,\,\,\,\,\, x=\pm 3\)
\(\textbf{3)}\) \(2x^2-6=26\)
The answer is \( x=\pm4 \)
\(\,\,\,\,\,\, 2x^2-6=26 \)
\(\,\,\,\,\,\, 2x^2=32 \)
\(\,\,\,\,\,\, x^2=16 \)
\(\,\,\,\,\,\, \sqrt{x^2}=\pm\sqrt{16}\)
\(\,\,\,\,\,\, x=\pm 4\)
\(\textbf{4)}\) \(\left(x-5\right)^2=25\)
The answer is \( x=0 \text{ or } x=10 \)
\(\,\,\,\,\,\, \left(x-5\right)^2=25 \)
\(\,\,\,\,\,\, \sqrt{\left(x-5\right)^2}=\pm\sqrt{25} \)
\(\,\,\,\,\,\, x-5=\pm5 \)
\(\,\,\,\,\,\, x=\pm5+5 \)
\(\,\,\,\,\,\, x=-5+5 \text{ or } x=5+5\)
\(\,\,\,\,\,\, x=0 \text{ or } x=10\)
\(\textbf{5)}\) \(3\left(x-2\right)^2+1=4\)
The answer is \( x=1 \text{ or } x=3 \)
\(\,\,\,\,\,\, 3\left(x-2\right)^2+1=4 \)
\(\,\,\,\,\,\, 3\left(x-2\right)^2=3 \)
\(\,\,\,\,\,\, \left(x-2\right)^2=1 \)
\(\,\,\,\,\,\, \sqrt{\left(x-2\right)^2}=\pm\sqrt{1} \)
\(\,\,\,\,\,\, x-2=\pm 1 \)
\(\,\,\,\,\,\, x=\pm 1 + 2 \)
\(\,\,\,\,\,\, x=-1+2 \text{ or } x= 1+2 \)
\(\,\,\,\,\,\, x=1 \text{ or } x= 3 \)
\(\textbf{6)}\) \(2\left(x+1\right)^2+3=11\)
The answer is \( x=1 \text{ or } x=-3 \)
\(\,\,\,\,\,\, 2\left(x+1\right)^2+3=11 \)
\(\,\,\,\,\,\, 2\left(x+1\right)^2=8 \)
\(\,\,\,\,\,\, \left(x+1\right)^2=4 \)
\(\,\,\,\,\,\, \sqrt{\left(x+1\right)^2}=\pm\sqrt{4} \)
\(\,\,\,\,\,\, x+1=\pm 2 \)
\(\,\,\,\,\,\, x=-2-1 \text{ or } x= 2-1 \)
\(\,\,\,\,\,\, x=-3 \text{ or } x=1 \)
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