Quadratic Equations?

Question

x = -16 plus/minus root 16^2 - 4 x 0 x 63 / 2

Can someone please explain to me how to get -12 and -8 from this equation. I can get to 'exact roots' but can't figure out the rest. Thanks! :o)

Answer 1

You can't have a 0 inside there...You would multiply it by a 1.

x^2 + 16x + 63; that x^2 is 1, not 0.

Therefore,

-16 +/- root(16^2 - 4(1)(63)) / 2 = -16 +/- root(4) / 2

which is -16 +/- 2 / 2, which simplifies to -8 +/- 1 (or -9 and -7)

Answer 2

x = -16 plus/minus root 16^2 - 4 x 0 x 63 / 2
x = [-16 +- SQRT(16^2 - 4(0)(63))]/2

As the formula for a quadratic equation, ax^2 + bx + c = 0, is:
x = [-b +- SQRT(b^2 - 4ac)]/2a,

I can see that b=16 and c=63 but is a=0 (from 4ac) or 1 (from 2a)? You must have substituted it wrongly.

If a=1, then the solution becomes:
x = [-16 +- SQRT(16^2 - 4(1)(63))]/2(1)
x = [-16 +- SQRT(4)]/2
x = [-16 +- 2]/2
x = -14/2 or -18/2
x = -7 or -9

Answer 3

I assume you mean

x = (-16 +/- root(16^2 - 4*0*63)) / 2.

The sum of the roots is going to be -16, which is inconsistent with roots of -12 and -8.

This obviously came from an attempt to solve a quadratic equation. What was this equation?

Answer 4

I gotta rewrite your expression: (please tell me if it's different)--

x = (-16 +- sqrt(16^2 - 4*0*63))/2

The second term in the root is zero, so the root is that of 16^2, or 4

This reduces to (-16 +- 4)/2 or -8 +- 2 = -10, -6

If the -16 isn't supposed to be divided by 2, the answer is

-16 +- 2 = -18, -14

If this is supposed to be the soulution to a quadratic eq, you're out of luck, because you have to divide by 2a, not just 2, and your a value is zero.

Answer 5

Think you have made an error....

the eqn you have written is x = -b+/-root(b^2-4ac)/2a

where a=coeff of x^2. b=coeff x and c=constant

this equation is used when you are unable to factorise a normal quadratic equation in the form ax^2 + bx + c

The way you have written it you have a= to 1 and also = to 0??
c=63 and b =16.

basically what I am saying is that a cannot = 0 and you have fucked up.

Mind you I have been taking acid for the last few hours and could be talking utter shite

Answer 6

You can't.

x = [-b +/- sqrt(b^2-4ac)]/2a

Given your original statement
a=0, b=16, c=63

This the equation for a line: y=16x+63 not for a quadratic function.

Or you wrote it wrong and a=1, b=16, c=63. Then

x = [-16 +/- sqrt(16^2-4*1*63)]/2
x = [-16 +/- sqrt(4)]/2
x = (-16 +/- 2)/2

x = -7 or -9

Answer 7

the initial equation was probably

x^2 +20x+96=0

using the quadratic formula, this gives

x= (-20 + or - sqrt(20^2 - 4*1*96))/2*1

= (-20 + or - sqrt(400-384))/2

= (-20 + or - sqrt(16) )/2 = (-24)/2 or(-16/2)

= -12 or -8

i hope that this helps

Answer 8

x = (-16 ± sqrt(16^2 - 4(0)(63)))/2

I think you have something wrong, for you have have a divisor of 2, "a" must be 1

x = (-16 ± sqrt(16^2 - 4(1)(63)))/(2(1))
x = (-16 ± sqrt(256 - 252))/2
x = (-16 ± sqrt(4))/2
x = (-16 ± 2)/2
x = -8 ± 1

x = -9 or -7

-------------------------------

(x + 12)(x + 8) = x^2 + 8x + 12x + 96 = x^2 + 20x + 96

so i think you have something wrong.

Answer 9

You made an error in the expression you wrote. If the 0 is a one you will get -7 and -9 (not the numbers you wrote).

Answer 10

there is no answer to this because 16 divided by zero is undefined. therefore you can't go any further.