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Solve \(\displaystyle x + 1 = \sqrt{7 x + 15}\).
Squaring both sides gives \(\displaystyle x^{2} + 2 x + 1 = 7 x + 15\). The equation is quadratic setting it equal to zero gives \(\displaystyle x^{2} - 5 x - 14 = 0\). Factoring gives \(\displaystyle (x - 7)(x + 2)=0\) so the possible solutions are \(\displaystyle x = 7\) and \(\displaystyle x = -2\). Checking the solution \(\displaystyle x = 7\) in the original equation gives \(\displaystyle 8 = 8\). The solution checks, so \(\displaystyle x = 7\) is a true solution. Checking the solution \(\displaystyle x = -2\) in the original equation gives \(\displaystyle -1 = 1\). The solution does no check, so \(\displaystyle x = -2\) is an extraneous solution.
\begin{question}Solve $x + 1 = \sqrt{7 x + 15}$.
\soln{10cm}{Squaring both sides gives $x^{2} + 2 x + 1 = 7 x + 15$. The equation is quadratic setting it equal to zero gives $x^{2} - 5 x - 14 = 0$. Factoring gives $(x - 7)(x + 2)=0$ so the possible solutions are $x = 7$ and $x = -2$. Checking the solution $x = 7$ in the original equation gives $8 = 8$. The solution checks, so $x = 7$ is a true solution. Checking the solution $x = -2$ in the original equation gives $-1 = 1$. The solution does no check, so $x = -2$ is an extraneous solution. }
\end{question}
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\begin{document}\begin{question}(10pts) The question goes here!
\soln{9cm}{The solution goes here.}
\end{question}\end{document}<p> <p>Solve <img class="equation_image" title=" \displaystyle x + 1 = \sqrt{7 x + 15} " src="/equation_images/%20%5Cdisplaystyle%20x%20%2B%201%20%3D%20%5Csqrt%7B7%20x%20%2B%2015%7D%20" alt="LaTeX: \displaystyle x + 1 = \sqrt{7 x + 15} " data-equation-content=" \displaystyle x + 1 = \sqrt{7 x + 15} " /> .</p> </p><p> <p>Squaring both sides gives <img class="equation_image" title=" \displaystyle x^{2} + 2 x + 1 = 7 x + 15 " src="/equation_images/%20%5Cdisplaystyle%20x%5E%7B2%7D%20%2B%202%20x%20%2B%201%20%3D%207%20x%20%2B%2015%20" alt="LaTeX: \displaystyle x^{2} + 2 x + 1 = 7 x + 15 " data-equation-content=" \displaystyle x^{2} + 2 x + 1 = 7 x + 15 " /> . The equation is quadratic setting it equal to zero gives <img class="equation_image" title=" \displaystyle x^{2} - 5 x - 14 = 0 " src="/equation_images/%20%5Cdisplaystyle%20x%5E%7B2%7D%20-%205%20x%20-%2014%20%3D%200%20" alt="LaTeX: \displaystyle x^{2} - 5 x - 14 = 0 " data-equation-content=" \displaystyle x^{2} - 5 x - 14 = 0 " /> . Factoring gives <img class="equation_image" title=" \displaystyle (x - 7)(x + 2)=0 " src="/equation_images/%20%5Cdisplaystyle%20%28x%20-%207%29%28x%20%2B%202%29%3D0%20" alt="LaTeX: \displaystyle (x - 7)(x + 2)=0 " data-equation-content=" \displaystyle (x - 7)(x + 2)=0 " /> so the possible solutions are <img class="equation_image" title=" \displaystyle x = 7 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%207%20" alt="LaTeX: \displaystyle x = 7 " data-equation-content=" \displaystyle x = 7 " /> and <img class="equation_image" title=" \displaystyle x = -2 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20-2%20" alt="LaTeX: \displaystyle x = -2 " data-equation-content=" \displaystyle x = -2 " /> . Checking the solution <img class="equation_image" title=" \displaystyle x = 7 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%207%20" alt="LaTeX: \displaystyle x = 7 " data-equation-content=" \displaystyle x = 7 " /> in the original equation gives <img class="equation_image" title=" \displaystyle 8 = 8 " src="/equation_images/%20%5Cdisplaystyle%208%20%3D%208%20" alt="LaTeX: \displaystyle 8 = 8 " data-equation-content=" \displaystyle 8 = 8 " /> . The solution checks, so <img class="equation_image" title=" \displaystyle x = 7 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%207%20" alt="LaTeX: \displaystyle x = 7 " data-equation-content=" \displaystyle x = 7 " /> is a true solution. Checking the solution <img class="equation_image" title=" \displaystyle x = -2 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20-2%20" alt="LaTeX: \displaystyle x = -2 " data-equation-content=" \displaystyle x = -2 " /> in the original equation gives <img class="equation_image" title=" \displaystyle -1 = 1 " src="/equation_images/%20%5Cdisplaystyle%20-1%20%3D%201%20" alt="LaTeX: \displaystyle -1 = 1 " data-equation-content=" \displaystyle -1 = 1 " /> . The solution does no check, so <img class="equation_image" title=" \displaystyle x = -2 " src="/equation_images/%20%5Cdisplaystyle%20x%20%3D%20-2%20" alt="LaTeX: \displaystyle x = -2 " data-equation-content=" \displaystyle x = -2 " /> is an extraneous solution. </p> </p>