Welcome! Today we're exploring the Product Formula and Special Conditions on Zeros — a powerful tool that turns word problems into instant equations.

The product formula is the partner of the sum formula, and it handles a different set of problems.

The most powerful application is translating verbal conditions into instant algebraic equations:

Verbal Condition	What It Means
"reciprocal zeros"	One zero is the flip of the other
"negatives of each other"	One zero is the opposite sign
"one zero is 0"	Exactly what it says

By the end of this lesson, you will build a translation dictionary that converts any verbal condition into a sum or product equation.

Verbal condition → Algebraic equation → Solved!

1. Reciprocal zeros mean product = 1

📋 Given Info

Let's look at a concept that comes up often in polynomial problems.

The word 'reciprocal' appears frequently. Here's the key translation:

If one zero is the reciprocal of the other, their product equals 1.

Why? If the zeros are $α$ and $\frac{1}{α}$ , then:

\alpha \times \frac{1}{\alpha} = 1

= 1

So 'reciprocal zeros' directly means product = 1.

✍️ Question

Question 📝

The zeros of $3 x^{2} + 8 x + k$ are reciprocals of each other.

Find $k$ .

Reciprocal zerosOne zero is the flip of the other means: if one zero is $α$ , the other is $\frac{1}{\alpha}$ The other zero is the reciprocal.

Their product is always: $α \times \frac{1}{α} =$ $1$ always!When you multiply reciprocals, you get 1 every time.

✍️ MCQ

Choose one

If the zeros of a quadratic are reciprocals of each other, what is their product?

✓

So whenever you see 'reciprocal zeros'The moment you see this in any problem, it directly translates to: product of zeros = 1That's the condition you write down and use.

Now let's apply this to the quadratic $3 x^{2} + 8 x + k$ .

Using the product formulaThe tool for multiplying zeros together: $product of zeros = \frac{c}{a} =$ $\frac{k}{3}$ The unknown k sits in the numerator.

✍️ MCQ

Choose one

We know the zeros are reciprocals. What should

\frac{k}{3}

be equal to?

✓

Since the zeros are reciprocalsThat's the whole trick, we set this equal to 1keyThat's your equation: $\frac{k}{3} = 1$

Solving: $k = 3$ .

2. Negatives of each other mean sum = 0

Let's look at another common verbal condition in polynomial problems.

When we're told that the zeros are negatives of each other, this means:

If one zero is $α$ , the other is $- α$

Their sum would be:

\alpha + (-\alpha) = 0

Key insight: Zeros that are negatives of each other always have a sum of zero.

✍️ Question

Problem 📝

The zeros of $2 x^{2} + k x + 6$ are negatives of each other.

Find the value of $k$ ?.

If the zeros are $\alpha$ and $-\alpha$ (negatives of each other):

Sum = $α + (- α) =$ $0$ key!When zeros are negatives of each other

✍️ MCQ

Choose one

For the polynomial

2x^2 + kx + 6

, what is the sum of zeros equal to (using the formula)?

✓

Now let's apply this to $2x^2 + kx + 6$ .

Here $a = 2$ , $b = k$ .

Using the sum formula: Sum $= - \frac{b}{a} =$ $-\frac{k}{2}$

Since the sum must equal 0This is what happens when zeros are negatives of each other: $- \frac{k}{2} = 0$

So $k = 0$ The value that makes the zeros negatives of each other.

✍️ MCQ

Choose one

If the zeros of

5x^2 + mx + 3

are negatives of each other, what is the value of

m

?

✓

This means the polynomial becomes $2 x^{2} + 6$ , which has no x-term.

Notice how the middle term completely disappearsWhen this disappears, zeros are negatives of each other!

Key Rule: When the zeros are negatives of each otherThis pattern means the x-coefficient is always zero, the x-coefficient is always 0Remember this for exams.

This is because the sum of zeros $= 0$ Sum equals negative b over a, and sum $= - \frac{b}{a}$ , so $b = 0$ keyThat's why there's no x-term.

✍️ MCQ

Choose one

A quadratic

ax^2 + bx + c

has zeros that are negatives of each other. Which coefficient must be zero?

✓

3. One zero is double the other

Let's look at a problem involving a special condition on the zeros of a quadratic.

The 'double' condition — where one zero is double the other — requires us to use both the sum and product formulas together.

This gives us two equations to work with:

The sum formula helps us find the individual zeros
The product formula then gives us the unknown coefficient

📋 Given Info

Setting up the problem:

If the zeros are $α$ and $2\alpha$ (one is double the other), then:

Sum of zeros: $\alpha + 2\alpha = 3\alpha$
Product of zeros: $\alpha \cdot 2\alpha = 2\alpha^2$

✍️ Question

Problem 📝

One zero of $x^2 - 8x + k$ is double the other.

Find $k$ ?.

When one zero is double the otherYour cue to use alpha and 2 alpha, let the zeros be $\alpha$ This setup works for any ratio and $2\alpha$ If one is triple, you'd use alpha and 3 alpha.

From the sum formulaGives you a simple equation in one variable: $\alpha + 2\alpha = 3\alpha$ Once you set this up, the sum formula works $= \frac{- b}{a} = \frac{- (- 8)}{1} =$ $8$ sumThat's the whole trick.

So $α = \frac{8}{3}$ . The other zero is $2 \times \frac{8}{3} = \frac{16}{3}$ .

✍️ MCQ

Choose one

We found

\alpha = \frac{8}{3}

and

2\alpha = \frac{16}{3}

. To find

k

, which formula should we use next?

✓

From the product formula: $α \times 2 α = 2 α^{2} = \frac{c}{a} = \frac{k}{1} =$ $k$ productOur second equation connecting alpha to k

Substituting $\alpha = \frac{8}{3}$ We plug this value into the product equation into our product expression:

k = 2\alpha^2 = 2 \times \left(\frac{8}{3}\right)^2 = 2 \times \frac{64}{9} = \frac{128}{9}

(Use both formulas together to find unknowns)

✍️ MCQ

Choose one

What is the value of

k

?

✓

VerificationHow you catch silly mistakes before the examiner does:

Let's check our answer by verifying both conditions.

SumAlways check both against the formulas = $\frac{8}{3} + \frac{16}{3} = \frac{24}{3} = 8$ . And $\frac{- b}{a} = 8$ . ✓ Correct.

ProductThe key habit — not just one = $\frac{8}{3} \times \frac{16}{3} = \frac{128}{9} = k$ . And $\frac{c}{a} = \frac{128}{9}$ . ✓ CorrectverifiedOnly then can you be confident your answer is correct.

Both conditions are satisfied, so k = $\frac{128}{9}$ is our final answer.

✍️ MCQ

Choose one

In this problem, we used both the sum and product formulas. Which formula directly gave us the value of

k

?

✓