Answer:
[tex]x=4[/tex]
Step-by-step explanation:
Given [tex]\displaystyle\\\sqrt{x+12}-\sqrt{2x+1}=1[/tex], start by squaring both sides to work towards isolating [tex]x[/tex]:
[tex]\displaystyle\\\left(\sqrt{x+12}-\sqrt{2x+1}\right)^2=\left(1\right)^2[/tex]
Recall [tex](a-b)^2=a^2-2ab+b^2[/tex] and [tex]\sqrt{a}\cdot \sqrt{b}=\sqrt{a\cdot b}[/tex]:
[tex]\displaystyle\\\left(\sqrt{x+12}-\sqrt{2x+1}\right)^2=\left(1\right)^2\\\implies x+12-2\sqrt{(x+12)(2x+1)}+2x+1=1[/tex]
Isolate the radical:
[tex]\displaystyle\\x+12-2\sqrt{(x+12)(2x+1)}+2x+1=1\\\implies -2\sqrt{(x+12)(2x+1)}=-3x-12\\\implies \sqrt{(x+12)(2x+1)}=\frac{-3x-12}{-2}[/tex]
Square both sides:
[tex]\displaystyle\\(x+12)(2x+1)=\left(\frac{-3x-12}{-2}\right)^2[/tex]
Expand using FOIL and [tex](a+b)^2=a^2+2ab+b^2[/tex]:
[tex]\displaystyle\\2x^2+25x+12=\frac{9}{4}x^2+18x+36[/tex]
Move everything to one side to get a quadratic:
[tex]\displaystyle-\frac{1}{4}x^2+7x-24=0[/tex]
Solving using the quadratic formula:
A quadratic in [tex]ax^2+bx+c[/tex] has real solutions [tex]\displaystyle x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]. In [tex]\displaystyle-\frac{1}{4}x^2+7x-24[/tex], assign values:
[tex]\displaystyle \\a=-\frac{1}{4}\\b=7\\c=-24[/tex]
Solving yields:
[tex]\displaystyle\\x=\frac{-7\pm \sqrt{7^2-4\left(-\frac{1}{4}\right)\left(-24\right)}}{2\left(-\frac{1}{4}\right)}\\\\x=\frac{-7\pm \sqrt{25}}{-\frac{1}{2}}\\\\\begin{cases}x=\frac{-7+5}{-0.5}=\frac{-2}{-0.5}=\boxed{4}\\x=\frac{-7-5}{-0.5}=\frac{-12}{-0.5}=24 \:(\text{Extraneous})\end{cases}[/tex]
Only [tex]x=4[/tex] works when plugged in the original equation. Therefore, [tex]x=24[/tex] is extraneous and the only solution is [tex]\boxed{x=4}[/tex]
