Review Questions Answer
- Has an undefined slope and passes through the point (4,3)
\[x=3\]
- Passes through the point (2,1) and is perpendicular to \(y= \frac{-2}{5}x + 3\)
\[\eqalign{-1&=&\frac{-2}{5}m_p \\ m_p&=& -\frac{-5}{2}=2.5\\}\]
- Add these complex numbers: \((3-2i)+(4-i)\)
\[\eqalign{(3-2i)+(4-i)\\ (3+4) - (2i +i)\\ 7-3i}\]
- Simplify: \(\sqrt{-4} + \sqrt{-16}\)
\[\eqalign{\sqrt{-4} + \sqrt{-16}\\ 2i + 4i\\ 6i\\}\]
- Multiply: \(5i(5-3i)\)
\[\eqalign{5i(5-3i)\\ 25i-15i^2\\ 15 + 25i\\ 5(3+5i)}\]
- Divide: \(\frac{4-i}{2+3i}\)
\[\eqalign{&\llap{ } 0.5\\ 4-i&\overline{)2+3i\quad}\\ &\underline{\ 2-0.5i }\\ &\qquad 3.5i\rlap{\ (remainder)}\\}\]
Answer: \(0.5 + \frac{3.5i}{4-i}\)
- Solve this quadratic equation and write the two complex roots in \(a+bi\) form: \(x^2-4x +7=0\)
\[\eqalign{x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\ \frac{-(-4)\pm\sqrt{(-4)^2-4\cdot1\cdot\ 7}}{2\cdot1}\\ \frac{4\pm\sqrt{16-28}}{2}\\ \frac{4\pm2\sqrt{4-7}}{2}\\ 2\pm\sqrt{-3}\\ 2\pm i\sqrt{3}\\}\]
- Solve: \((3x-1)^2 - 1= 24\)
\[\eqalign{(3x-1)^2 - 1&=& 24\\ (3x-1)^2 &=& 25\\ \sqrt{(3x-1)^2} &=& \sqrt{25}\\ 3x -1 &=& 5\\ 3x &=& 6\\ x&=&2}\]
- Solve: \(x^2 -6x = 13\)
\[\eqalign{x^2 -6x &=& 13\\ x^2 -6x -13 &=& 0\\ x &=& \frac{6 \pm \sqrt{6^2-4(-13)}}{2}\\ &=& \frac{6 \pm \sqrt{36+52}}{2}\\ &=& \frac{6 \pm \sqrt{88}}{2}\\ &=& 3\pm \frac{9.38}{2}=3\pm 4.69=7.69, -1.69\\}\]
- Solve: \(4x^2 -4x - 1 = 0\)
\[\eqalign{4x^2 -4x - 1 &=& 0\\ x&=&\frac{4 \pm \sqrt{16 + 16}}{2\cdot 4}\\ &=&\frac{1\pm\sqrt{2}}{2}\\}\]
- Solve: \(\sqrt{x - 7} = x- 7\)
\[\eqalign{\sqrt{x - 7} &=& x- 7\\ \sqrt{x - 7} &=& \left(\sqrt{x- 7}\right)^2\\ 1 &=& \sqrt{x- 7}\\ 1^2 &=& (\sqrt{x- 7})^2\\ 8 = x\\ }\]
- Solve: \(2 + \sqrt{12 - 2x} = x\)
\[\eqalign{2 + \sqrt{12 - 2x} = x}\]
- Solve: (x-1)^ = 9
\[\eqalign{2 + \sqrt{12 - 2x} &=& x\\ \sqrt{12 - 2x} &=& x-2\\ 12 -2x &=& x^2 -4x + 2\\ 0 &=& x^2 -2x - 10\\ x &=& \frac{2 \pm \sqrt{2^2 -4\cdot -10\cdot 1}}{2}\\ &=& \frac{2 \pm \sqrt{16}}{2}\\ &=& 1\pm 4= -3, \ 5\\ }\]