Question

Solve the equation using the quadratic formula. 4x^2+12x+58=0

Answer 1

`~~~~~~~~~~~~\textit{quadratic formula} \\\\ \stackrel{\stackrel{\textit{\small a}}{\downarrow }}{4}x^2\stackrel{\stackrel{\textit{\small b}}{\downarrow }}{+12}x\stackrel{\stackrel{\textit{\small c}}{\downarrow }}{+58}=0 \qquad \qquad x= \cfrac{ - b \pm \sqrt { b^2 -4 a c}}{2 a}`

`x= \cfrac{ - (12) \pm \sqrt { (12)^2 -4(4)(58)}}{2(4)} \implies x = \cfrac{ -12 \pm \sqrt { 144 -928}}{ 8 } \\\\\\ x= \cfrac{ -12 \pm \sqrt { -784 }}{ 8 }\implies x= \cfrac{ -12 \pm \sqrt { (-1)(28^2) }}{ 8 }\implies x=\cfrac{ -12 \pm 28\sqrt { -1 }}{ 8 } \\\\\\ x=\cfrac{ -3 \pm 7\sqrt { -1 }}{ 2 }\implies x=\cfrac{ -3 \pm 7i}{ 2 }\implies x= \begin{cases} ( -3 + 7i)/( 2 )\\\\ ( -3 - 7i)/( 2 ) \end{cases}`

whenever we get imaginary or "complex" solutions, is another way of saying the function has no real solutions or it really never touches the x-axis.