FIND THE ANSWERS

# How to solve 4(6y+3)>28 and also how to solve 2x^2-2x-40<0?

### Answer this question

• How to solve 4(6y+3)>28 and also how to solve 2x^2-2x-40<0?

## Answers

Answer #1 | 22/05 2017 23:14
Question 1 6y + 3 > 7 6y > 4 y > 2/3 Question 2 x² - x - 20 < 0 [ x - 5 ] [ x + 4 ] < 0 _______________-4__________5________ x - 5_____-ve_________-ve____0___+ve____ x + 4____-ve_____0___+ve________+ve___ product__+ve____ 0___-ve____0___+ve____ solution set is {x : -4 < x < 5 }
Answer #2 | 22/05 2017 22:48
I'll give you some general theory first. If you have a continuous function (which polynomials are, and [thus] are connected as well) we note that the only way for a function to change sign is to pass through 0. Thus, if we inspect the sign of the function between the x-coordinates where the function hits zero, we can (in terms of positivity and negativity) determine the behavior of the polynomial. So, we can get the second one to be x^2 - x - 20 < 0, i.e. (x - 5)(x + 4) < 0. Thus, we hit zero at x = 5 and x = -4. Then, since the sign doesn't change between these points, we can just pick a point in each interval and see where negativity holds. Inspection tells me only when x is in (-4, 5) do we have negativity. For the first, just solve, 6y + 3 > 7, 6y > 4, y > 2/3.

### Possible answer

Anonymous607264
Login to your account
Create new account

Discover Questions