Question

Using the quadratic formula to solve x2 + 20 = 2x, what are the values of x?

Answer 1

The values of x are:

1+√19i

1-√19i

Step-by-step explanation:

Given equation is:

$x^2+20=2x\\x^2-2x+20=0\\Comparing\ with\ the\ standard\ form\ of\ quadratic\ equation\\ax^2+bx+c = 0\\a=1\\b=-2\\c=20Using Quadratic formula\\x = \frac{-b+\sqrt{b^2-4ac}}{2a}\ \ \ , x = \frac{-b-\sqrt{b^2-4ac}}{2a}\\x = \frac{-(-2)+\sqrt{(-2)^2-4(1)(20)}}{2(1)}\ \ \ , x = \frac{-(-2)-\sqrt{(-2)^2-4(1)(20)}}{2(1)}\\x = \frac{2+\sqrt{4-80)}}{2}\ \ \ , x = \frac{2-\sqrt{4-80)}}{2}\\x = \frac{2+\sqrt{-76)}}{2}\ \ \ , x = \frac{2-\sqrt{-76)}}{2}$

$x = \frac{2+\sqrt{-19*4}}{2}\ \ \ , x = \frac{2-\sqrt{-19*4}}{2}\\x = \frac{2+\sqrt{-19*4}}{2}\ \ \ , x = \frac{2-\sqrt{-19*4}}{2}\\x = \frac{2+2\sqrt{-19}}{2}\ \ \ , x = \frac{2-2\sqrt{-19}}{2}\\x = \frac{2(1+\sqrt{-19})}{2}\ \ \ , x = \frac{2(1-\sqrt{-19})}{2}\\x = 1+\sqrt{-19}\ \ \ , x = 1-\sqrt{-19}\\\sqrt{-19} = \sqrt{19}*\sqrt{-1} \\As\ we\ know => \sqrt{-1} =i\\\sqrt{-19} = \sqrt{19}i\\So,\\x= 1+\sqrt{19}i\ \ \ \ , x=1-\sqrt{19}i$

The values of x are:

1+√19i

1-√19i

Keywords: Quadratic equation, value of variable

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