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00:05
12 Apr 2026

Computing Symmetric Expressions Without Finding Zeros

Welcome! Today we're tackling a clever technique — Computing Symmetric Expressions Without Finding Zeros.

Some problems ask you to compute an expression involving the zeros — like:

  • 1α+1β\frac{1}{\alpha} + \frac{1}{\beta}
  • αβ+βα\frac{\alpha}{\beta} + \frac{\beta}{\alpha}

...without actually finding the zeros.

The trick: Rewrite the expression in terms of (α+β)(\alpha + \beta) and α⋅β\alpha \cdot \beta, which you can read directly from the coefficients.

A few identities from Class 8 algebra make this possible.

What you needWhere it comes from
α+β\alpha + \betaCoefficients
α⋅β\alpha \cdot \betaCoefficients
Algebraic identitiesClass 8

These problems are designed to be solved this way — the zeros are often irrationalmessy!, making direct computation very messy.

1. The identity for 1/alpha + 1/beta

Here's a clever technique: sometimes problems ask you to compute expressions involving the zeros of a polynomial — like 1α+1β\frac{1}{\alpha} + \frac{1}{\beta} — without actually finding the zeros themselves.

The trick? Rewrite the expression in terms of the sum and product of zeros, which you can read directly from the coefficients!

📋 Given Info

The simplest symmetric expression identity:

1α+1β=α+βα⋅β=SP\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha \cdot \beta} = \frac{S}{P}

This is just common-denominator addition!

It converts a reciprocal-sum into a ratio of:

  • SS = sum of zeros (α+β)(\alpha + \beta)
  • PP = product of zeros (α⋅β)(\alpha \cdot \beta)
✍️ Question

Your turn! 🧮

If α\alpha and β\beta are zeros of 2x2+3x−52x^2 + 3x - 5, find:

1α+1β\frac{1}{\alpha} + \frac{1}{\beta}
find this

Without finding the individual zeros.

The identity: 1α+1β=α+βα⋅β\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha \cdot \beta}Use this for 1/α + 1/β questions

This is just adding fractionsAdding fractions with common denominator with a common denominatorThe common denominator is alpha times beta: 1α+1β=β+αα⋅β\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\beta + \alpha}{\alpha \cdot \beta}

✍️ MCQ
Choose one
For the quadratic 2x2+3x−52x^2 + 3x - 5, what are the values of SS (sum of zeros) and PP (product of zeros)?
✓

For 2x2+3x−52x^2 + 3x - 5:

α+β=−ba\alpha + \beta = -\frac{b}{a}Sum of roots from coefficients =−32= -\frac{3}{2}

α⋅β=ca\alpha \cdot \beta = \frac{c}{a}Product of roots from coefficients =−52= -\frac{5}{2}

So:

1α+1β=α+βα⋅β</pen>=<penactions="circle"circle−color="blue"circle−annotation="plugin"narrationText="pluginthevaluesyoualreadycalculated"commentary="Substitutethesumandproductvalues">−3/2−5/2\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha \cdot \beta}</pen> = <pen actions="circle" circle-color="blue" circle-annotation="plug in" narrationText="plug in the values you already calculated" commentary="Substitute the sum and product values">\frac{-3/2}{-5/2}
(Key identity: reciprocals become sum over product)

✍️ MCQ
Choose one
When we divide −3/2−5/2\frac{-3/2}{-5/2}, what operation do we actually perform?
✓

Simplifying: −3/2−5/2=\frac{-3/2}{-5/2} = −32×2−5\frac{-3}{2} \times \frac{2}{-5}Standard fraction division technique == 35\frac{3}{5}Got the answer using only sum and product

✍️ MCQ
Choose one
If a different quadratic has α+β=4\alpha + \beta = 4 and α⋅β=−2\alpha \cdot \beta = -2, what is 1α+1β\frac{1}{\alpha} + \frac{1}{\beta}?
✓

Dividing fractions: −3/2−5/2=(−32)×(−25)=610=35\frac{-3/2}{-5/2} = \left(-\frac{3}{2}\right) \times \left(-\frac{2}{5}\right) = \frac{6}{10} = \frac{3}{5} flip and multiplyThe standard method for dividing fractions 2s cancel negative × negative = positiveWatch your signs carefully here 3/5

We never found α\alpha or β\beta individually. We went straight from coefficients to the answer. That is the power of the identity approach.The identity approach lets us skip finding roots never found α or βNo need to solve for individual roots straight from coefficients to answerGo directly from coefficients to the final result clean algebra

✍️ MCQ
Choose one
For a quadratic with sum of zeros S=4S = 4 and product of zeros P=−2P = -2, what is 1α+1β\frac{1}{\alpha} + \frac{1}{\beta}?
✓

2. The identity for alpha^2 + beta^2

Let's look at a powerful technique! 🎯

The Identity:

α2+β2=(α+β)2−2αβ\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta

This lets us rewrite a sum of squares entirely in terms of:

  • SS (sum of zeros)
  • PP (product of zeros)

Where does it come from?

Expanding (α+β)2(\alpha + \beta)^2:

(α+β)2=α2+2αβ+β2(\alpha + \beta)^2 = \alpha^2 + 2\alpha\beta + \beta^2

Rearranging: α2+β2=(α+β)2−2αβ\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta

This identity is the building block for more complex expressions like αβ+βα\frac{\alpha}{\beta} + \frac{\beta}{\alpha}advanced.

✍️ Question

Your Turn ✏️

If α\alpha and β\beta are zeros of 6x2+x−26x^2 + x - 2, compute α2+β2\alpha^2 + \beta^2.

The key identity we need is:

α2+β2=(α+β)2−2αβ\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta
(Use this identity for sum of squares problems)

This lets us compute α2+β2\alpha^2 + \beta^2 using only the sum and product of the zeros — no need to find α\alpha and β\beta individuallyJust read off sum and product from coefficients.

✍️ MCQ
Choose one
For the quadratic 6x2+x−26x^2 + x - 2, what is the value of α+β\alpha + \beta?
✓

For the polynomial 6x2+x−26x^2 + x - 2, we first extract the sum and product of zeros using Vieta's formulasFind sum and product directly from coefficients.

Here a=6a = 6, b=1b = 1, c=−2c = -2

Sum of zeros:The negative sign is easy to forget

α+β=−ba=−16\alpha + \beta = -\frac{b}{a} = -\frac{1}{6}
(Remember this pattern)

Product of zeros:No negative sign here, just straight division

αβ=ca=−26=−13\alpha\beta = \frac{c}{a} = \frac{-2}{6} = -\frac{1}{3}
(Straight division of coefficients)

✍️ MCQ
Choose one
What is (α+β)2(\alpha + \beta)^2 when α+β=−16\alpha + \beta = -\frac{1}{6}?
✓

Now we apply the identityThe identity you need to memorize step by step:

Step 1: (α+β)2(\alpha + \beta)^2Always compute S squared first =(−16)2== \left(-\frac{1}{6}\right)^2 = 136\frac{1}{36}S²(S squared computed separately)

Step 2: 2αβ2\alpha\beta =2×(−13)== 2 \times \left(-\frac{1}{3}\right) = −23-\frac{2}{3}2P

✍️ MCQ
Choose one
We have (α+β)2=136(\alpha + \beta)^2 = \frac{1}{36} and 2αβ=−232\alpha\beta = -\frac{2}{3}. What operation gives us α2+β2\alpha^2 + \beta^2?
✓

Step 3: α2+β2=136−(−23)=\alpha^2 + \beta^2 = \frac{1}{36} - \left(-\frac{2}{3}\right) = 136+23\frac{1}{36} + \frac{2}{3}Minus a negative becomes plus

Step 4: Convert 23\frac{2}{3} to 36ths: 23=2436\frac{2}{3} = \frac{24}{36}Standard fraction addition method

Step 5: 136+2436=\frac{1}{36} + \frac{24}{36} = 2536\frac{25}{36}Answer!Got the answer using sum and product directly

✍️ Yes/No
Yes or No?
Could we compute α2+β2\alpha^2 + \beta^2 by first finding α\alpha and β\beta using the quadratic formula, then squaring and adding?
✓

3. Computing alpha/beta + beta/alpha

Now let's tackle a slightly more complex expression: αβ+βα\frac{\alpha}{\beta} + \frac{\beta}{\alpha}

This one is interesting because it combines both previous identities.

Key insight:

αβ+βα=α2+β2α⋅β\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha \cdot \beta}
key formula

Notice:

  • The numerator uses the sum-of-squares identity you just computed
  • The denominator is simply αβ\alpha\beta — the product of zeros
✍️ Question

Your Turn 🧮

For the same polynomial 6x2+x−26x^2 + x - 2, compute:

αβ+βα\frac{\alpha}{\beta} + \frac{\beta}{\alpha}

Hint: Use the identity αβ+βα=α2+β2α⋅β\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha \cdot \beta} and the values you computed earlier.

To compute αβ+βα\frac{\alpha}{\beta} + \frac{\beta}{\alpha}, we use this identityYour go-to move for this expression type:

αβ+βα=α2+β2α⋅β\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha \cdot \beta}
(Numerator is sum of squares, denominator is product)

✍️ MCQ
Choose one
Why can we rewrite αβ+βα\frac{\alpha}{\beta} + \frac{\beta}{\alpha} as α2+β2α⋅β\frac{\alpha^2 + \beta^2}{\alpha \cdot \beta}?
✓

We already know from our previous work that α2+β2=2536\alpha^2 + \beta^2 = \frac{25}{36} and α⋅β=−13\alpha \cdot \beta = -\frac{1}{3}.

✍️ MCQ
Choose one
Using α2+β2=2536\alpha^2 + \beta^2 = \frac{25}{36} and αβ=−13\alpha\beta = -\frac{1}{3}, what is αβ+βα\frac{\alpha}{\beta} + \frac{\beta}{\alpha}?
✓

Now we divide:

25/36−1/3\frac{25/36}{-1/3}

Dividing by a fraction = multiplying by its reciprocalWhen dividing by a fraction, flip and multiply:

2536×−31=−7536\frac{25}{36} \times \frac{-3}{1} = \frac{-75}{36}

Simplify by dividing numerator and denominator by 3Look for common factors to reduce:

−7536=−2512\frac{-75}{36} = \frac{-25}{12}

So αβ+βα=−2512\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = -\frac{25}{12}The sum and product were enough

✍️ MCQ
Choose one
In this problem, we computed αβ+βα\frac{\alpha}{\beta} + \frac{\beta}{\alpha} by dividing α2+β2\alpha^2 + \beta^2 by αβ\alpha\beta. Which values did we use?
✓

Notice the chain of reasoning: the coefficient formula gives us SS and PPS and P come straight from coefficients, no solving needed, the identity (α+β)2−2αβ(\alpha + \beta)^2 - 2\alpha\betaUse this identity to get the sum of squares gives us α2+β2\alpha^2 + \beta^2, then division gives the final answerDivision by P completes the calculation.

✍️ Yes/No
Yes or No?
In this problem, we computed αβ+βα\frac{\alpha}{\beta} + \frac{\beta}{\alpha} for 6x2+x−26x^2 + x - 2. Did we ever find the actual values of α\alpha and β\beta?
✓

No individual zeros were computed at any point.The main takeaway from this method

This is the power of symmetric expressionsWork only with sum and product — we work entirely with SS and PP from Vieta's formulasUsing S and P from Vieta's formulas, never needing to solve the quadraticNo solving required at all!