How do I find all real and complex zeros of $x^{2} - 4 x + 7$ $x^2-4x+7$ ?

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How do I find all real and complex zeros of $x^{2} - 4 x + 7$ $x^2-4x+7$ ?

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Answer 1 · 2015-10-31T10:53:52

Use the quadratic formula to find zeros occur for:

$x = 2 \pm i \sqrt{3}$ $x = 2+-i sqrt(3)$

Explanation:

The quickest way is probably to use the quadratic formula.

$x^{2} - 4 x + 7$ $x^2-4x+7$ is in the form $a x^{2} + b x + c$ $ax^2+bx+c$ , with $a = 1$ $a=1$ , $b = - 4$ $b=-4$ and $c = 7$ $c=7$ .

So its zeros occur for:

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} = \frac{- (- 4) \pm \sqrt{{(- 4)}^{2} - (4 \times 1 \times 7)}}{2 \times 1}$ $x = (-b+-sqrt(b^2-4ac))/(2a) = (-(-4)+-sqrt((-4)^2-(4xx1xx7)))/(2xx1)$

$= \frac{4 \pm \sqrt{16 - 28}}{2} = \frac{4 \pm \sqrt{- 12}}{2} = \frac{4 \pm i \sqrt{12}}{2}$ $=(4+-sqrt(16-28))/2 = (4+-sqrt(-12))/2 = (4+-i sqrt(12))/2$

$= \frac{4 \pm i \cdot 2 \sqrt{3}}{2} = 2 \pm i \sqrt{3}$ $=(4+-i*2sqrt(3))/2 = 2+-i sqrt(3)$

Alternatively, you can complete the square as follows:

$0 = x^{2} - 4 x + 7 = x^{2} - 4 x + 4 + 3 = {(x - 2)}^{2} + 3$ $0 = x^2-4x+7 = x^2-4x+4+3 = (x-2)^2+3$

Subtract $3$ $3$ from both ends to get:

${(x - 2)}^{2} = - 3$ $(x-2)^2 = -3$

So:

$x - 2 = \pm \sqrt{- 3} = \pm i \sqrt{3}$ $x-2 = +-sqrt(-3) = +-i sqrt(3)$

Add $2$ $2$ to both sides to get:

$x = 2 \pm i \sqrt{3}$ $x = 2 +-i sqrt(3)$

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How do I find all real and complex zeros of $x^{2} - 4 x + 7$ $x^2-4x+7$ ?

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How do I find all real and complex zeros of $x^{2} - 4 x + 7$ $x^2-4x+7$ ?