Dear Greg Mat family, could you help to untie the knots entangling me in this simple inequality question? sort of frustrated that I am still stuck at rudimentary question like this…
I managed to solve it up to (x - 6)(x + 5) < 0, where x < 6 and X < -5, but the answer is -5 < x < 6. Specifically for X + 5 < 0, do we need to flip the sign when we deduct -5 from both sides of the inequality? I thought that’s only applicable if we divide or multiply a negative value?
Thank you in advance.
The solution works like this: what happens to (x - 6) and (x + 5) when -5 < x < 6? There is no flipping needed as once we have gotten the (x - 6)(x + 5) < 0 point, we are asking the question “what happens to (x - 6)(x + 5) when -5 < x < 6?” You can show that it must always be negative.
Ah i see. So if I put anything > -5, the answer returns a positive value which contradicts the inequality sign of the original equation. During the exam, due to limited time, I wonder if I can heuristically assume that a quadratic will always have a +ve and -ve sign? Ie the constant we bring to the RHS which becomes negative will have to flip the inequality sign?
More of “x > 6” or “x < 5”
I’m not sure what you mean - give an example.
Hi thanks for making time to teach me on this.
What I meant was since it’s a quadratic, it will create two inequalities. For example, from the factoring we did, (X - 6) ( X + 5) > 0. In this scenario, if I have to minus from both sides of the inequality, I should flip the inequality. Therefore,
(X - 6) > 0
X > 6
(X + 5) > 0
X < -5
Would this work or merely a fluke? Thank you in advance.
I don’t see the inequality flipping though. And no, you don’t flip the inequality when adding/subtracting negative numbers. You do it only for multiplication/division.
Thanks. I jump headlong into solving the inequality. Seems like the best way to validate inequality is to test out the range. It only make sense when I try validating it by inputting x = -6 and x = -4. Yet to build the intuition to solve it fast.
