Question

What are the solutions of 4x2 + x = –3?

Answer 1

Get all the terms on one side. So,
\[4x^2 + x + 3 =0\] Then you need to factorise.
You're either going to have (x + a)(4x + b) or (2x+c)(2x+d)
You need to find a combination where ab or cd = 3 and where 4ax+bx or 2cx+2dx = x

Answer 2

Seeing as the factors of 3 are only 1 and 3, a and b or c and d must be 1 and 3.

Answer 3

okay

Answer 4

i get that so far

Answer 5

If you put that into the second equation you get
(2x+3)(2x+1)= 4x^2 + 8x + 3 so it's not that.
So its either (4x+1)(x+3) or (4x+3)(x+1)

Answer 6

i think it's (4x + 1)(x + 3)

Answer 7

Wait, they don't work either, are you sure it's not 4x^2 + x = 3?

Answer 8

it must be then

Answer 9

i know that the answer is going to be an imaginary number and a radical all over an integer

Answer 10

i got -1 but i can be wrong

Answer 11

Right well put it into the quadratic formula.\[x =(- b +- \sqrt{b^2-4ac} )/2a\]

Answer 12

b = 1, a = 4 and c = 3.

Answer 13

im working it out on paper to see if i can get it

Answer 14

Ok :P This equation doesn't have any real roots so they'll both be imaginary.

Answer 15

I got -1 + or - i√47 all over 8

Answer 16

is that right?

Answer 17

No. You can't have 1 real root and 1 imaginary. Either 2 real or 2 imaginary.

Answer 18

Oh right!! Yeah that's right.

Answer 19

The 'or' confused me.

Answer 20

wait so whats the final answer?

Answer 21

\[\frac{ -1 + \sqrt{47}i }{ 8 }\]

Answer 22

Or\[\frac{ -1 - \sqrt{47}i }{ 8 }\]

Answer 23

i being root -1 and therefore imaginary.

Answer 24

kk