Solve f(x) = x^2 + 2x + 7 in simplest a + bi form

so, what does f(x) = ? ... -95, 14, 21.1, -2, 0, ... :?:
 
Trig ace said:
=o
sorry
Click to expand...

\(\displaystyle \L\ ax^2 + bx + c = 0, a \not=\ 0\)

There's a formula for these:

\(\displaystyle \L\ x = \frac{-b + i sqrt{4ac - b^2}}{2a}\, x = \frac{-b - i sqrt{4ac - b^2}}{2a}\\)
 
Morson, wouldn't haveing the i on there cancel out and make the negative square root * i make the whole thing a real number?
 
Pay better attention, Random. After noticing that b^2 - 4ac is negative, what does that make 4ac - b^2?
 
What is the point of a special quadratic formula for quadratics with imaginary solutions anyways? If you use the ordinary one, and get something like SQRT(-10), you pull out an i, and you've got SQRT(10).
Memorizing more formulas for special cases doesn't really help improve your math...
 
merlin2007 said:
What is the point of a special quadratic formula for quadratics with imaginary solutions anyways? If you use the ordinary one, and get something like SQRT(-10), you pull out an i, and you've got SQRT(10).
Memorizing more formulas for special cases doesn't really help improve your math...
Click to expand...

Very good point, Merlin.

The quadratic formula works all the time.....whether the solutions it produces are rational, real, equal, unequal, or imaginary.

No "special versions" needed.
 
Who's to say which formula is more appropriate to use? Both formulas "work all the time." The formula I posted above does not require rewriting square roots of negative numbers, unless \(\displaystyle b^2 > 4ac\).

The standard quadratic formula does not require rewriting square roots of negative numbers, unless \(\displaystyle b^2 < 4ac\).

It also depends on context. I suggested the formula as the OP is evidently studying, or was studying, solving equations over the complex numbers. However, students who work with real numbers only would fancy the standard formula more. Dependant on context.
 
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