# Solving Quadratic Equations: Why are Values Negative?

• Gringo123
In summary: In the above equation, a, b, and c represent constants and x is the variable. First, multiply the a term and the c term. Then find two numbers whose sum is b and whose product is ac. After this you can factor by grouping after rewriting the equation using the two numbers you obtained in the previous step.

#### Gringo123

Here is a quadratic equation along with its solution. There's only 1 thing I don't understand. In the brackets how do we know that the values are - 4 and - 5 as opposed to + 4 and + 5. Why are they both negative when the quadratic contains both a negative and a positive expression?

x2 - 9 + 20 = 0

(x - 4) (x - 5) = 0

x = 4 or 5

Gringo123 said:
Here is a quadratic equation along with its solution. There's only 1 thing I don't understand. In the brackets how do we know that the values are - 4 and - 5 as opposed to + 4 and + 5. Why are they both negative when the quadratic contains both a negative and a positive expression?

x2 - 9 + 20 = 0

(x - 4) (x - 5) = 0

x = 4 or 5

Let me ask you a question.
If we have ab=0 ,
what does it suggest?
[at least either a or b has to be 0]

What happens when you multiply two negatives vs. adding two negatives?

I'll assume your original quadratic should read $$x^2 - 9x + 20 = 0$$.

When you have a quadratic with leading coefficient equal to 1, and integer coefficients, there is a simple process to follow to check whether the polynomial will or will not factor with integer constants.
Step 1: Write down all pairs of integers with product equal to the constant term.

Step 2: Look in your list for a pair whose SUM equals the middle coefficient. These are the appropriate choices. (IF there is no such pair, the given quadratic won't factor this way)

For yours, the constant is 20. There are two possible choices: $$4 \text{ and } 5$$ and $$-4 \text{ and } -5$$. Since the second pair sum to $$-9$$, these are the choices, and you know

$$x^2 - 9x + 20 = (x-4)(x-5)$$

As another example, consider factoring

$$x^2 - 6x - 27 = 0$$

Step 1: Look for integer pairs that multipy to $$-27$$. There are two choices: $$3 \text{ and } -9$$, and $$-3 \text{ and } 9$$. The first pair has sum $$-6$$, so those are your choices.

$$x^2 - 6x- 27 = 0 \Rightarrow (x-9)(x+3) = 0$$

so the solutions are $$x = 9 \text{ and } x = -3$$.

Finally, consider $$x^2 - 11x + 8 = 0$$. Sets of integers with a product of $$8$$ are $$1 \text{ and } 8, 2 \text{ and } 4, -1 \text{ and } -8, -2 \text{ and } -4$$. None of these pairs sum to $$-11$$, so this polynomial doesn't factor using integers.

It is very important to remember that this method works ONLY when the leading coefficient (the coefficient of $$x^2$$) equals 1.

Gringo123 said:
Here is a quadratic equation along with its solution. There's only 1 thing I don't understand. In the brackets how do we know that the values are - 4 and - 5 as opposed to + 4 and + 5. Why are they both negative when the quadratic contains both a negative and a positive expression?

x2 - 9 + 20 = 0

(x - 4) (x - 5) = 0

x = 4 or 5
[itex](x- a)(x- b)= x^2+ (-a-b)x+ (-a)(-b)

(-a-b)= -(a+b) but (-a)(-b)= ab- the sum of two negative numbers is negative but the product of two negative numbers is positive.

-4-5= -9, (-4)(-5)= +20.

Thanks a lot everyone. It's a lot clearer now.

Thanks to help from various people on this site I have learned how to solve trinomial quadratics when the coefficient of x2 is 1. However, I've now moved on to this type of quadratic. It seems that a different approach is needed:

2x2 - 7x - 15 = 0

I know from looking at the answers, that the factors required are as follows:
(2x + 3) (x - 5) = 0
But how did we arrive at + 3 and - 5? They don't add to make -7.

Gringo123 said:
Thanks to help from various people on this site I have learned how to solve trinomial quadratics when the coefficient of x2 is 1. However, I've now moved on to this type of quadratic. It seems that a different approach is needed:

2x2 - 7x - 15 = 0

I know from looking at the answers, that the factors required are as follows:
(2x + 3) (x - 5) = 0
But how did we arrive at + 3 and - 5? They don't add to make -7.

For quadratic equations where the coefficient of x2 is not 1, I like to use the "AC method."

ax2 + bx + c = 0

In the above equation, a, b, and c represent constants and x is the variable. First, multiply the a term and the c term. Then find two numbers whose sum is b and whose product is ac. After this you can factor by grouping after rewriting the equation using the two numbers you obtained in the previous step.

m + n = b
(m)(n) = (a)(c)

(ax2 + mx) + (nx + c) = 0

## 1. Why do quadratic equations have negative solutions?

Quadratic equations can have negative solutions because they involve finding the roots or solutions to an equation in the form of ax^2 + bx + c = 0. Since the quadratic equation involves squaring the variable, there is a possibility for both positive and negative solutions.

## 2. Can a quadratic equation have only one negative solution?

Yes, a quadratic equation can have only one negative solution. This occurs when the equation's discriminant, or b^2 - 4ac, is equal to 0. In this case, the equation has a double root, or one solution repeated twice, which can be either positive or negative.

## 3. How can we determine the sign of the solutions to a quadratic equation?

The sign of the solutions to a quadratic equation can be determined by looking at the equation's discriminant. If the discriminant is positive, the equation will have two distinct real solutions, one positive and one negative. If the discriminant is negative, the equation will have two complex solutions with imaginary components. If the discriminant is equal to 0, the equation will have one real solution, which can be either positive or negative.

## 4. What does it mean when a quadratic equation has no negative solutions?

When a quadratic equation has no negative solutions, it means that the equation has two distinct real solutions, both of which are positive. This can occur when the discriminant is positive or when the equation has no real solutions, but two complex solutions with positive real components.

## 5. How does the sign of the coefficients in a quadratic equation affect the solutions?

The sign of the coefficients in a quadratic equation does not affect the solutions. The solutions to a quadratic equation are only affected by the values of the coefficients, not their signs. However, the signs of the coefficients can determine the concavity of the graph of the equation, which can affect the number of real solutions.