Question

How do you solve `4x^{2} + 16x = - 7`?

Answer 1

`-2 +- (5sqrt2)/4`

Explanation:

`y = 4x^2 + 16x + 7 = 0`
Solve it by the improved quadratic formula in graphic form:
`D = d^2 = b^2 - 4ac = 256 - 56 = 200`--> `d = +- 10sqrt2`
There are 2 real roots:
`x = -b/(2a) +- d/(2a) = -16/8 +- (10sqrt2)/8 = -2 +- (5sqrt2)/4`

Answer 2

x = `-1/2, -7/2`

Explanation:

(1) here `4x^2 +16x+7 = 0`
By using the quadratic Equation formula we can write
x = `[-b+-sqrt{b^2-4ac}]/[2a]` where b=16, a=4 and c=7.
so, x = `[-16+-sqrt{16^2-4.4.7}]/[2.4]`
or, x = `[-16+-sqrt144]/8`
or, x = [-16+12]/8, [-16-12]/8
or, x = `-1/2, -7/2`

OR (2)
`4X^2+16X+7 = 0` Multiply both sides by 2, we get
`8X^2+32X+14 = 0`
or, `8X^2+4X+28X+14 = 0`
or, 4X(2X+1)+14(2X+1) = 0
or, (4X+14)(2X+1)= 0
or, X = `-1/2, -7/2`