[size=150]Clare solves the quadratic equation [math]4x^2+12x+58=0[/math], but when she checks her answer, she realizes she made a mistake. Explain what Clare's mistake was.[/size][br][br][math]x=\frac{\text{-}12\pm\sqrt{12^2-4\cdot4\cdot58}}{2\cdot4}[/math][br][math]x=\frac{\text{-}12\pm\sqrt{144-928}}{8}[/math][br][math]x=\frac{\text{-}12\pm\sqrt{\text{-}784}}{8}[/math][br][math]x=\frac{\text{-}12\pm28i}{8}[/math][br][math]x=\text{-}1.5\pm28i[/math]

[size=150]Write in the form [math]a+bi[/math], where [math]a[/math] and [math]b[/math] are real numbers:[/size][br][br][math]\frac{5\pm\sqrt{\text{-}4}}{3}[/math]

[math]\frac{10\pm\sqrt{\text{-}16}}{2}[/math]

[math]\frac{\text{-}3\pm\sqrt{\text{-}144}}{6}[/math]

[size=150]For the first equation, she writes [math]x=\frac{4\pm\sqrt{16-72}}{12}[/math][br]For the second equation, she writes [math]x=\frac{8\pm\sqrt{64-24}}{6}[/math][/size][br][br]Which equation(s) will have real solutions? Which equation(s) will have non-real solutions? Explain how you know.

[math]x^2=25[/math]

[math]x^3=27[/math]

[math]x^2=12[/math]

[math]x^3=12[/math]

[size=150]Kiran is solving the equation [math]\sqrt{x+2}-5=11[/math] and decides to start by squaring both sides. Which equation results if Kiran squares both sides as his first step?[/size]