Question

X^2+8x+1<0 Inequality questions What will be the answer ? Plz tell

Answer 1

`-4-sqrt(15) < x < -4+sqrt(15)`

Explanation:

Complete the square:
`x^2+8x+1<0`
`(x+4)^2-15<0`
`(x+4)^2<15`
`|x+4| < sqrt(15)`

If `x+4>=0`, then `x<-4+sqrt(15)`.
If `x+4<0`, then `-x-4 < sqrt(15)rArrx > -4-sqrt(15)`

So we have two ranges for `x`:
`-4<=x<-4+sqrt(15)` and `-4-sqrt(15) < x < -4`.
We can combine these to make one range:
`-4-sqrt(15) < x < -4+sqrt(15)`

Numerically, to three significant figures:
`-7.87 < x < -0.127`

Answer 2

`(-4 - sqrt15, -4 + sqrt15)`

Explanation:

`f(x) = x^2 + 8x + 1 < 0`
First, solve the quadratic equation f(x) = 0, to find the 2 end-points (critical points).
`D = d^2 = b^2 - 4ac = 64 - 4 = 60` --> `d = +- 2sqrt15`
There are 2 real roots:
`x = -b/(2a) +- d/(2a) = - 8/2 +- 2sqrt15/2 = -4 +- sqrt15`
`x1 = -4 - sqrt15`, and `x2 = - 4 + sqrt15)`.
The graph of f(x) is a upward parabola (a > 0). Between the 2 real roots (x1 , x2), the graph is below the x-axis --> f (x) < 0.
The answer is the open interval:
`(-4 - sqrt15 , -4 + sqrt15)`