Question

How do you solve `2b^{2} + 14= - 11b`?

Answer 1

`b =-2 or b= -7/2`

Explanation:

When you see the `x^2` term, you know it is a quadratic equation.

Make it equal to 0.

`2b^2 +11b +14 = 0 " "larr` try to find factors.

Find factors of 2 and 14 which ADD (because of PLUS 14) to give 11.
The signs will all be positive.

Find factors and combine them by cross multiplying to find pairs of factors which add to 11.

`color(white)(xxx)2" "14`

`color(white)(xxx)2" "7 rarr 1xx7 = 7`
`color(white)(xxx)1" "2rarr 2 xx2 = ul4`
`color(white)(xxxxxxxxxxxxxxxx)11`

The top row is the first bracket (2 and 7)
the second row is the second bracket (1 and 2)

`2b^2 +11b +14 = 0`

`(2b+7)(b+2)=0`

The product of two factors is 0, one of them MUST be 0.

Each factor could be equal to 0.

if `2b+7 = 0" "color(white)(xxxxxxx)if b+2 = 0`

`rarr 2b = -7" "color(white)(xxxxxxx) rarrb = -2`

`rarr b = -7/2`