Question

Solve the equation in the complex number system. x^2-12x+40=0

Answer 1

solving the equation $x^2-12x+40=0$ we get $x=6+2i\,\,and\,\,x=6-2i$

Step-by-step explanation:

Solve the equation:

$x^2-12x+40=0$

We can solve using quadratic Formula:

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

a= 1, b = -12, c = 40

Putting values:

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\x=\frac{-(-12)\pm\sqrt{(-12)^2-4(1)(40)}}{2(1)}\\x=\frac{12\pm\sqrt{144-160}}{2}\\x=\frac{12\pm\sqrt{-16}}{2}\\We\,\,know\,\,\sqrt{-1}=i\\ x=\frac{12\pm4i}{2}\\ x=\frac{12+4i}{2}\,\,and\,\, x=\frac{12-4i}{2}\\ x=\frac{4(3+i)}{2}\,\,and\,\, x=\frac{4(3-i)}{2}\\x=2(3+i)\,\,and\,\,x=2(3-i)\\x=6+2i\,\,and\,\,x=6-2i$

So, solving the equation $x^2-12x+40=0$ we get $x=6+2i\,\,and\,\,x=6-2i$

Keywords: Solving quadratic equation

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