Question

Solving Quadratic Equations:

what is the square root of -7x²=-448 ?

show work please :/

Answer 1

there is nothing to work out, the index \[\sqrt[n]{x}\]n being the index, is even and the radicand, x, is negative, so that is not a real number

Answer 2

well what about taking the square root of x²=-2x+33 can that be worked out?

Answer 3

that would be \[x=\sqrt{-2x+33}\]

Answer 4

hmmm, how about solving algebraically. So
1. -7x^2 = -448
2. (-7x^2)/-7 = -448/-7 //divide each side by -7
3. x^2 = 64
4. Squrt(x^2) = Squrt(64) //take the square root of both sides to get rid of the exponent
5. x = 8

Answer 5

or does it have to be done as a quadratic?

Answer 6

hmm.. my assignment says solving quadratic functions and to take the square root of the number I listed above.

Answer 7

@Confucious where do I go from there?

Answer 8

hmm, I think in the case of solving as a quadratic:
1. -7x^2 = -448
2. -7x + 0x +448 = 0
3. So in the form of ax^2 + bx +c = 0, a = -7, b = 0, c = 448.

Answer 9

oh that's it, you leave it like that

Answer 10

heh also I was wrong btw. x^2 = 64 means x=8 or x=-8

Answer 11

so now plug in a=-7, b = 0, and c=448 into http://en.wikipedia.org/wiki/Quadratic_equation - both variations (plus and minus), to solve for x

Answer 12

For your first problem, do as chris said \(-7x²=-448 \implies x^2=64 \implies x=\pm 8\).

Answer 13

are you doing nonreal numbers? do you know about \[i\]?

Answer 14

For the second problem \(x²=-2x+33 \iff x^2+2x-33=0\), you need to use the quadratic formula \(\large x={-b \pm \sqrt{b^2-4ac} \over 2a}\), with a=1, b=2 and c=-33.

Answer 15

@Chris you are sort of confusing me..

I got this part:
1. -7x^2 = -448
2. -7x + 0x +448 = 0
3. So in the form of ax^2 + bx +c = 0, a = -7, b = 0, c = 448.
but not your second set of directions?

Answer 16

@Confucious: The problem has solutions in real numbers.

Answer 17

wait my answer was wrong, use the quadratic formula to solve it, you can do it that way

Answer 18

Like Mr. Math said (if I'm reading the question correctly) - if your goal is to solve for x, the form of ax^2 + bx + c = 0 allows you to use the quadratic equation Mr. Math posted to solve for both solutions of x

Answer 19

just by plugging in the numbers a, b, and c

Answer 20

@Confucious so that's it? You don't take the square root because you can't? I'm unsure about the unreal numbers.

Answer 21

@Mandy17, use the quadratic formula like Mr. Math said and you can solve it

Answer 22

@Mr. Math, which problem are you talking about?

Answer 23

mandy - read the first two sections of this page - http://en.wikipedia.org/wiki/Quadratic_equation
Basically if you can write a formula in the form of : ax^2 + bx + c = 0, where x is unknown and a, b, and c are known, you can use the quadratic formula to solve for x - the unknown.

Answer 24

the idea is, given an equation - use algebra to put zero on one side of it, and see if you have something that resembles ax^2+bx + c

Answer 25

and my answer I come up with from the quadratic formula is my answer to: taking the square root?

Answer 26

in the first problem, we did:
1. -7x²=-448
2. -7x^2 +448 = -448 + 448 //add 448 to both sides so that zero is on the right hand side
3. -7x^2 + 448 = 0
4. Now look to see if you have something in the form of ax^2 + bx + c = 0. You do!
5. a = -7 (look above), since there is nothing that's just multiplied by x ("bx"), you still have a "b", but it's zero, and c is 448.
6. i.e. -7x^2 + 0x + 448 = 0, and since this is the format of a quadratic equation, we can use a = -7, b = 0, c = 448 , and stick them into the formula that Mr. Math put in.

Answer 27

you should get -8 and +8 (since really the quadratic formula always has you do two cases - "b +/-" means "b plus or minus" - so you have to plug in the numbers twice. One where you add to b, one where you subtract.

Answer 28

@Christ the final total outcome of the quadratic formula is -8 and +8?
or is there more I need to do.