[math]w^2=\text{-}367[/math]
[math](z+5)^2=\text{-}367[/math]
[size=150]Kiran was using the quadratic formula to solve the equation [math]x^2-12x+41=0[/math]. [br]He wrote this:[br][center][math]x=\frac{12\pm\sqrt{144-164}}{2}[/math][/center]Then he noticed that the number inside the square root is negative and said, “This equation doesn’t have any solutions.”[br][/size][br]Do you agree with Kiran? Explain your reasoning.
Write [math]\sqrt{\text{-}20}[/math] as an imaginary number.
[size=150]Although imaginary numbers let us describe complex solutions to quadratic equations, they were actually discovered and accepted because they could help us find real solutions to equations with polynomials of degree 3. In the 16th century, mathematicians discovered a cubic formula for solving equations of degree 3, but to use it they sometimes had to work with complex numbers. Let’s see an example where this comes up.[/size][br][br]To find a solution to the equation [math]x^3-px-q=0[/math] the cubic formula would first tell us to find a complex number, [math]z[/math], which is [math]\frac{q}{2}+i\sqrt{(\frac{p}{3})^3-(\frac{q}{2})^2}[/math]. Find [math]z[/math] when our equation is [math]x^3-15x-4=0[/math].[br]
The next step is to find a complex number [math]w[/math] so that [math]w^3=z[/math]. Show that [math]w=2+i[/math] works for the [math]z[/math] we found in step 1.[br]
If we write [math]w=a+b[/math]i where [math]a[/math] and [math]b[/math] are real numbers, the solutions to our equation are [math]2a[/math], [math]-a+b\sqrt{3}[/math], and [math]-a-b\sqrt{3}[/math]. What are the three solutions to our equation [math]x^3-15x-a=0[/math]?[br]