Question

How do you solve `8x^2+4=-33x`?

Answer 1

The solutions are:
`x = color(blue)(-1/8`

`x= color(blue)( - 4`

Explanation:

`8x^2 + 4 = -33x`

`8x^2 + 4 + 33x = 0`

`8x^2 + 33x + 4 = 0`

The equation is of the form `color(blue)(ax^2+bx+c=0` where:

`a=8, b=33, c=4`

The Discriminant is given by:

`Delta=b^2-4*a*c`

`= (33)^2-(4* 8 * 4)`

`= 1089 -128= 961`

The solutions are found using the formula
`x=(-b+-sqrtDelta)/(2*a)`

`=(-33+-sqrt961)/(2*8)`

`=(-33+-31)/(16)`

`x =(-33+31)/(16) = -2/16 = color(blue)(-1/8`

`x =(-33-31)/(16) = -64/16 = color(blue)( - 4`

Answer 2

1/8 and 4

Explanation:

`y = 8x^2 + 33x + 4 = 0`
Use the new Transforming Method (Google and Yahoo Search).
Transformed equation: `y' = x^2 + 33x + 32`.
Since a - b + c = 0, use shortcut. Two real roots of y' are: -1 and (-c/a = -32).
Back to y, divide these real roots by (a = 8) to get the 2 real roots of y --> : `x1 = -1/8` and `x2 = -32/8 = -4.`