Question

What are the solutions of x2 - 2x +17 - 0?

Answer 1

Answer:

option B

Step-by-step explanation:

$x^2 - 2x + 17 = 0 \ , \ [ \ where \ a = 1 , \ b = - 2 \ c = 17 \ ]$

$x = \frac{- b \ \pm \sqrt{b^2 - 4ac}}{2a}\\\\x = \frac{2 \ \pm \sqrt{2^2 - (4 \times 1 \times 17)}}{ 2 \times 1}\\\\x = \frac{2 \ \pm \sqrt{4 - 68}}{ 2 }\\\\x = \frac{2 \ \pm \sqrt{-64}}{ 2 }\\\\x = \frac{2 \ \pm \sqrt{8^2 \times -1}}{ 2 }\\\\x = \frac{2 \ \pm \ 8\sqrt{ -1}}{ 2 }$              

$x = \frac{2 \ \pm\ 8 i}{ 2 }$                               $[ \ where \ i = \ \sqrt{i} \ ]$

$x = 1 \ \pm \ 4i\\\\x = 1 + 4i , \ x = 1 - 4i$

Answer 2

Answer:

O B.  $x=1+4i$ or $x=1-4i$

Step-by-step explanation:

Use the Quadratic Formula for the Quadratic Equation!

Quadratic Equation: $x^{2} -2x+17=0$

Quadratic Formula: $x= \frac{-b±\sqrt{b^2-4ac} }{2a}$

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Now set up the values like this!

$ax^{2} +bx+c$

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$x=\frac{-b±\sqrt{-2^2-(4(0)(17)} }{2(0)}$

$x=\frac{-b±\sqrt{4-(1)} }{0}$

x = 1 - 4i

x = 1 + 4i