How do you solve and write the following in interval notation: $x^{2} - 2 x + 4 > 0$ $x^2-2x+4>0$ ?

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How do you solve and write the following in interval notation: $x^{2} - 2 x + 4 > 0$ $x^2-2x+4>0$ ?

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Answer 1 · 2018-05-07T03:19:43

inequality always true.

Explanation:

$f (x) = x^{2} - 2 x + 4 > 0$ $f(x) = x^2 - 2x + 4 > 0$
$D = d^{2} = b^{2} - 4 a c = 4 - 16 = - 12 < 0$ $D = d^2 = b^2 - 4ac = 4 - 16 = - 12 < 0$
Because D < 0, there are no real roots. The graph is an upward parabola (a > 0). The graph stays completely above the x-axis.
The inequality is always true.

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How do you solve and write the following in interval notation: $x^{2} - 2 x + 4 > 0$ $x^2-2x+4>0$ ?

1 Answer

Explanation:

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How do you solve and write the following in interval notation: x2−2x+4>0x^2-2x+4>0?

1 Answer

Explanation:

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How do you solve and write the following in interval notation: $x^{2} - 2 x + 4 > 0$ $x^2-2x+4>0$ ?