When a quadratic equation is always negative

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I looked this up and seen something that was beyond my A-Level Maths course.

In class we are doing the discriminant and sketching quadratic graphs, so it is nothing advanced. My teacher completed the square to prove the quadratic:

$$ 2x^2 + 8x + 9 $$

is always positive for all real values of $x$. So it was:

$$2(x + 2)^2 + 1$$

And her notes have the $2$ multiplying the $2$ in the bracket saying "always positive", and pointing to the $1$ saying always positive. I just don't understand how that proves it, and what if it was negative?

Any help appreciated. :)

asked Oct 14, 2015 at 21:29

Tiernan WatsonTiernan Watson

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$(x + 2)^2 \ge 0$ because it is a square. So $2(x + 2)^2 + 1 \ge 1 > 0$.

As $2(x+2)^2 + 1 = 2x^2 + 8x + 9$, $2x^2 + 8x + 9 > 0$

In general, not all quadratics will be entirely positive or entirely negative but you can always convert $ax^2 + bx + c = a(x^2 + b/ax + b^2/4a^2) + c - b^2/a = a(x + b/2a)^2 + (c - b^2/a)$ The term squared will always be non-negative. If a and (c - b^2/a) are both positive or are both negative the quadratic will be either always positive or negative. If they are not both then the quadratic will be be positive for some values and negative for others.

answered Oct 14, 2015 at 21:47

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whatever thing squared will always be positive. think about it, however small of a negative number x can be, after adding 2 to it, will still be vastly negative. however, when any negative number is squared it becomes negative - the negative of a negative is always positive.

answered Sep 24, 2017 at 19:11

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The square of a real number is always positive (doesn't have to be strictly positive, but still positive).

$2(x+2)^2 + 1$ is always positive because it is the sum of two positive numbers. $2(x+2)^2$ is the product of two positive numbers so it is positive. 1 is positive because it is a strictly positive number.

$2x^2+8x+9 = 2(x+2)^2 + 1 \geq 1 > 0$ so $2x^2+8x+9$ is not only always positive, but always strictly positive for all real $x$.

answered Dec 4, 2021 at 18:23

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Video transcript

We're asked to solve the quadratic equation, negative 3x squared plus 10x minus 3 is equal to 0. And it's already written in standard form. And there's many ways to solve this. But in particular, all solve it using the quadratic formula. So let me just rewrite it. We have negative 3x squared plus 10x minus 3 is equal to 0. And actually, I'll solve it twice using the quadratic formula to show you that as long as we manipulated this in the valid way, the quadratic formula will give us the exact same roots or the exact same solutions to this equation. So in this form right over here, what are our ABCs? Let's just remind ourselves what the quadratic formula even is actually. That's a good place to start. The quadratic formula tells us that if we have a quadratic equation in the form ax squared plus bx plus c is equal to 0, so in standard form, then the roots of this are x are equal to negative b plus or minus the square root of b squared minus 4ac, all of that over 2a. And this is derived from completing the square in a general way. So it's no magic here, and I've derived it in other videos. But this is the quadratic formula. This is actually giving you two solutions, because you have the positive square root here and the negative square root. So let's apply it here in the case where-- in this case, a is equal to negative 3, b is equal to 10, and c is equal to negative 3. So applying the quadratic formula right here, we get our solutions to be x is equal to negative b. b is 10. So negative b is negative 10 plus or minus the square root of b squared. b is 10. So b squared is 100 minus 4 times a times c. So minus 4 times negative 3 times negative 3. Let me just write it down. Minus 4 times negative 3 times negative 3. All of that's under the radical sign. And then all of that is over 2a. So 2 times a is negative 6. So this is going to be equal to negative 10 plus or minus the square root of 100 minus-- negative 3 times negative 3 is positive 9. Positive 9 times 4 is positive 36. We have a minus sign out here. So minus 36. All of that over negative 6. This is equal to 100 minus 36 is 64. So negative 10 plus or minus the square root of 64. All of that over negative 6. The principal square root of 64 is 8. But we're taking the positive and negative square root. So this is negative 10 plus or minus 8 over negative 6. So if we take the positive version, we say x could be equal to-- negative 10 plus 8 is negative 2 over negative 6. So that was taking the plus version. That's this right over here. And negative 2 over negative 6 is equal to 1/3. If we take the negative square root, negative 10 minus 8-- So let's take negative 10 minus 8. That would be x is equal to-- negative 10 minus 8 is negative 18. And that's going to be over negative 6. Negative 18 divided by negative 6 is positive 3. So the two roots for this quadratic equation are positive 1/3 and positive 3. And I want to show you the we'll get the same answer, even if we manipulate this. Some people might not like the fact that our first coefficient here is a negative 3. Maybe they want a positive 3. So to get rid of that negative 3, they can multiply both sides of this equation times negative 1. And then if you did that, you would get 3x squared minus 10x plus 3 is equal to 0 times negative 1, which is still equal to 0. So in this case, a is equal to 3, b is equal to negative 10, and c is equal to 3 again. And we could apply the quadratic formula. We get x is equal to negative b. b is negative 10. So negative negative 10 is positive 10, plus or minus the square root of b squared, which is negative 10 squared, which is 100, minus 4 times a times c. a times c is 9 times 4 is 36. So minus 36. All of that over 2 times a. All of that over 6. So this is equal to 10 plus or minus the square root of 64, or really that's just going to be 8. All of that over 6. If we add 8 here, we get 10 plus 8 is 18 over 6. We get x could be equal to 3. Or if we take the negative square root or the negative 8 here, 10 minus 8 is 2. 2 over 6 is 1/3. So once again, you get the exact same solutions.

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