when $\displaystyle c=-7 $and $\displaystyle d=4$

$\displaystyle \sqrt{d} - dc^2$

$\displaystyle \sqrt{4} - 4 x (-7)^2$

$\displaystyle 2 - 4 x (-7)^2 = -194$

this right?

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- Aug 7th 2009, 11:28 AMADYevaluation
when $\displaystyle c=-7 $and $\displaystyle d=4$

$\displaystyle \sqrt{d} - dc^2$

$\displaystyle \sqrt{4} - 4 x (-7)^2$

$\displaystyle 2 - 4 x (-7)^2 = -194$

this right? - Aug 7th 2009, 11:39 AMAmer
- Aug 7th 2009, 11:42 AMADY
So are you saying there are two answers?

- Aug 7th 2009, 11:44 AMAmer
- Aug 7th 2009, 11:45 AMADY
Ok so i'm not sure what to do now then?

- Aug 7th 2009, 11:52 AMAmer
did you know that

$\displaystyle (-2)^2=4$

$\displaystyle (-4)^2=16$

$\displaystyle (-7)^2=49$

so

$\displaystyle \sqrt{4}=\mp 2 $

$\displaystyle \sqrt{16}=\mp 4 $

$\displaystyle \sqrt{49}=\mp 7 $

in general

$\displaystyle \sqrt{a^2}=\mp a $ a is real number

in you question

$\displaystyle \sqrt{4}-4(-7)^2$ have two solutions since $\displaystyle \sqrt{4}=2 $ and $\displaystyle \sqrt{4}=-2$

first one

$\displaystyle 2-4(49)=2-196=-194$

second one

$\displaystyle -2-4(49)=-2-196=-198$ - Aug 7th 2009, 12:07 PMADY
Does the same apply to this formula

$\displaystyle a = -7, b = 4$

$\displaystyle a - b(a + b)$

$\displaystyle (-7) - 4((-7)+4) = 5$ - Aug 7th 2009, 12:10 PMAmer
- Aug 7th 2009, 12:20 PMstapel
One might

*say*that the evaluation has two answers, but this would be incorrect. (Blush)

While one*solves*"x^2 = 4" for the two solution values, -2 and +2, "the" value of the square root of 4, sqrt[4], is*defined*to be the principal (that is, the positive real) value: sqrt[4] = 2.

So no, there will most definitely*not*be two answers to this one evaluation exercise! (Surprised)

$\displaystyle \sqrt{d}\, -\, dc^2\, =\,\sqrt{4}\, -\, (4)(-7)^2$

. . .$\displaystyle =\, 2\, -\,4(49)\, =\, 2\, -\, 196\, =\, -194$

And the above is the*only*answer! (Wink)