x2 + 10x -56 < 0. i need help and fast, im very confused and i think i did something wrong

Answer 1

Don’t worry I have help for you !

x² + 10x - 56 < 0

(x + 14)(x -4) < 0

This breaks up our number line into three regions (-∝, -14) , (-14, 4) ,(4,∞). The points -14, 4 are the points that x =0. The values will be greater than zero or less than zero in each region. Let’s test a number in each region to see if it is < 0

Choosing -20 in the interval(-∞,-14) gives -6×-24 >0

Choosing 0 in the interval (-14, 4) gives 14×-4 < 0

Choosing 6 in the interval (4, ∞) gives 20 × 2 > 0

So (x + 14)(x -4) < 0 on the interval (-14, 4)

My final answer is (-14,4).

Answer 2

To find what x's will yield an x² + 10x - 56 value that is less than zero, let's consider the graph of y = x² + 10x - 56. I'm sure you area aware this is a parabola, To find the details, let's factor it into y = (x + 14)(x - 4) meaning it crosses the x-axis (where y = 0) at x = -14 and at x = 4. Those will be the boundaries for the solution (those are the x values that result in x² + 10x - 56 being equal to zero). Since the parabola opens up, the portion < 0 is the portion between the two x-intercepts.

So the solution is -14 < x < 4