The Key Relationship

Here is one of the most useful results in the whole chapter. For a quadratic polynomial

p(x)=ax2+bx+c,a≠0p(x) = ax^2 + bx + c, \quad a \neq 0

let its two zeroes be α\alpha (alpha) and β\beta (beta). Then:

Sum of zeroes=α+β=−ba=−coefficient of xcoefficient of x2\text{Sum of zeroes} = \alpha + \beta = -\frac{b}{a} = -\frac{\text{coefficient of } x}{\text{coefficient of } x^2}

Product of zeroes=αβ=ca=constant termcoefficient of x2\text{Product of zeroes} = \alpha\beta = \frac{c}{a} = \frac{\text{constant term}}{\text{coefficient of } x^2}

Key Point: You can find the sum and product of the zeroes without solving the polynomial — just read off aa, bb, cc. This saves enormous time in the exam.

[Board Important] Memorise the signs: sum is −b/a-b/a (note the minus), product is +c/a+c/a. Mixing up the sign on the sum is the most common error.

Why the Relationship Works

Let's see why this is true — understanding it helps you remember it.

If α\alpha and β\beta are the zeroes of ax2+bx+cax^2 + bx + c, we can write the polynomial in factored form:

ax2+bx+c=a(x−α)(x−β)ax^2 + bx + c = a(x - \alpha)(x - \beta)

Expanding the right side:

a(x−α)(x−β)=a[x2−(α+β)x+αβ]=ax2−a(α+β)x+aαβa(x - \alpha)(x - \beta) = a\left[x^2 - (\alpha + \beta)x + \alpha\beta\right] = ax^2 - a(\alpha+\beta)x + a\alpha\beta

Now compare coefficients with ax2+bx+cax^2 + bx + c:

  • Coefficient of xx: −a(α+β)=b⇒α+β=−ba-a(\alpha + \beta) = b \Rightarrow \alpha + \beta = -\dfrac{b}{a}.
  • Constant term: aαβ=c⇒αβ=caa\alpha\beta = c \Rightarrow \alpha\beta = \dfrac{c}{a}.

Key Point: The relationships come straight from expanding the factored form and matching coefficients. This is the same idea used later for cubics.

[Board Important] Sometimes a 2–3 mark question asks you to derive these relations. Showing this expansion earns full marks.

Using the Relationship — Verification

A very common exam task: find the zeroes of a quadratic, then verify the sum and product relations.

Worked outline: p(x)=x2+7x+10p(x) = x^2 + 7x + 10

  1. Find zeroes by factorising: x2+7x+10=(x+2)(x+5)x^2 + 7x + 10 = (x + 2)(x + 5), so α=−2\alpha = -2, β=−5\beta = -5.
  2. Sum check: α+β=−2+(−5)=−7\alpha + \beta = -2 + (-5) = -7. And −ba=−71=−7-\dfrac{b}{a} = -\dfrac{7}{1} = -7. ✓
  3. Product check: αβ=(−2)(−5)=10\alpha\beta = (-2)(-5) = 10. And ca=101=10\dfrac{c}{a} = \dfrac{10}{1} = 10. ✓

Key Point: Always present both the computed value (from the actual zeroes) and the formula value (from aa, bb, cc), and show they agree.

[Board Important] Even if you make a small factorisation slip, the verification structure (state formula, compute, compare) earns method marks.

Handy Identities with α\alpha and β\beta

Many questions give you a polynomial and ask for an expression in its zeroes without finding them individually. These identities turn the question into 'sum' and 'product' which you already know.

  • α2+β2=(α+β)2−2αβ\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta
  • 1α+1β=α+βαβ\dfrac{1}{\alpha} + \dfrac{1}{\beta} = \dfrac{\alpha + \beta}{\alpha\beta}
  • (α−β)2=(α+β)2−4αβ(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta
  • α3+β3=(α+β)3−3αβ(α+β)\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta)

Key Point: Express the required quantity in terms of (α+β)(\alpha + \beta) and αβ\alpha\beta, then substitute −b/a-b/a and c/ac/a. You rarely need the actual zeroes.

[Board Important] α2+β2=(α+β)2−2αβ\alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta is asked very often. Memorise it — it is the workhorse identity of this section.

Solved Examples

Example 1: Sum and product without solving

For p(x)=2x2−8x+6p(x) = 2x^2 - 8x + 6, find the sum and product of the zeroes.

Solution:

  1. Here a=2a = 2, b=−8b = -8, c=6c = 6.
  2. Sum =−ba=−−82=4= -\dfrac{b}{a} = -\dfrac{-8}{2} = 4.
  3. Product =ca=62=3= \dfrac{c}{a} = \dfrac{6}{2} = 3.

Final Answer: Sum =4= 4, Product =3= 3.

Takeaway: Read aa, bb, cc and apply the formulae — no solving needed.

Example 2: Find zeroes and verify

Find the zeroes of p(x)=x2+7x+10p(x) = x^2 + 7x + 10 and verify the relationship with the coefficients.

Solution:

  1. Factorise: (x+2)(x+5)=0⇒x=−2,−5(x + 2)(x + 5) = 0 \Rightarrow x = -2, -5.
  2. Sum: −2+(−5)=−7=−71=−ba-2 + (-5) = -7 = -\dfrac{7}{1} = -\dfrac{b}{a}. ✓
  3. Product: (−2)(−5)=10=101=ca(-2)(-5) = 10 = \dfrac{10}{1} = \dfrac{c}{a}. ✓

Final Answer: Zeroes −2,−5-2, -5; relations verified.

Takeaway: Compare computed sum/product with −b/a-b/a and c/ac/a.

Example 3: Zeroes of x2−3x^2 - 3

Find the zeroes of p(x)=x2−3p(x) = x^2 - 3 and verify the relations.

Solution:

  1. x2−3=0⇒x2=3⇒x=3,−3x^2 - 3 = 0 \Rightarrow x^2 = 3 \Rightarrow x = \sqrt{3}, -\sqrt{3}.
  2. Here a=1a = 1, b=0b = 0, c=−3c = -3.
  3. Sum: 3+(−3)=0=−01\sqrt{3} + (-\sqrt{3}) = 0 = -\dfrac{0}{1}. ✓
  4. Product: 3×(−3)=−3=−31\sqrt{3} \times (-\sqrt{3}) = -3 = \dfrac{-3}{1}. ✓

Final Answer: Zeroes 3,−3\sqrt{3}, -\sqrt{3}; verified.

Takeaway: When b=0b = 0, the zeroes are equal and opposite, so their sum is 0.

Example 4: Zeroes of 6x2−7x−36x^2 - 7x - 3

Find the zeroes of p(x)=6x2−7x−3p(x) = 6x^2 - 7x - 3 and verify the relationship.

Solution:

  1. Split the middle term: 6x2−9x+2x−3=3x(2x−3)+1(2x−3)=(3x+1)(2x−3)6x^2 - 9x + 2x - 3 = 3x(2x - 3) + 1(2x - 3) = (3x + 1)(2x - 3).
  2. Zeroes: x=−13x = -\dfrac{1}{3} and x=32x = \dfrac{3}{2}.
  3. Sum =−13+32=−2+96=76=−−76=−ba= -\dfrac{1}{3} + \dfrac{3}{2} = \dfrac{-2 + 9}{6} = \dfrac{7}{6} = -\dfrac{-7}{6} = -\dfrac{b}{a}. ✓
  4. Product =−13×32=−12=−36=ca= -\dfrac{1}{3} \times \dfrac{3}{2} = -\dfrac{1}{2} = \dfrac{-3}{6} = \dfrac{c}{a}. ✓

Final Answer: Zeroes −13,32-\dfrac{1}{3}, \dfrac{3}{2}; verified.

Takeaway: Splitting the middle term factorises a quadratic with a≠1a \neq 1.

Example 5: Find α2+β2\alpha^2 + \beta^2

If α\alpha and β\beta are the zeroes of x2−5x+6x^2 - 5x + 6, find α2+β2\alpha^2 + \beta^2.

Solution:

  1. Sum =5= 5, Product =6= 6 (from −b/a-b/a and c/ac/a).
  2. Use α2+β2=(α+β)2−2αβ\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta.
  3. =52−2(6)=25−12=13= 5^2 - 2(6) = 25 - 12 = 13.

Final Answer: α2+β2=13\alpha^2 + \beta^2 = 13.

Takeaway: Convert to sum and product, then substitute — no need to find α\alpha, β\beta.

Example 6: Find 1α+1β\dfrac{1}{\alpha} + \dfrac{1}{\beta}

If α,β\alpha, \beta are zeroes of 2x2−4x+12x^2 - 4x + 1, find 1α+1β\dfrac{1}{\alpha} + \dfrac{1}{\beta}.

Solution:

  1. Sum =−−42=2= -\dfrac{-4}{2} = 2; Product =12= \dfrac{1}{2}.
  2. 1α+1β=α+βαβ=21/2=4\dfrac{1}{\alpha} + \dfrac{1}{\beta} = \dfrac{\alpha + \beta}{\alpha\beta} = \dfrac{2}{1/2} = 4.

Final Answer: 1α+1β=4\dfrac{1}{\alpha} + \dfrac{1}{\beta} = 4.

Takeaway: 1α+1β=sumproduct\dfrac{1}{\alpha} + \dfrac{1}{\beta} = \dfrac{\text{sum}}{\text{product}}.

Example 7: Find a coefficient from a zero

If one zero of x2−kx+6x^2 - kx + 6 is 2, find kk and the other zero.

Solution:

  1. Product of zeroes =ca=61=6= \dfrac{c}{a} = \dfrac{6}{1} = 6. If one zero is 2, the other is 6/2=36/2 = 3.
  2. Sum of zeroes =2+3=5=−−k1=k= 2 + 3 = 5 = -\dfrac{-k}{1} = k.

Final Answer: k=5k = 5, other zero =3= 3.

Takeaway: Use the product to find the missing zero, then the sum to find kk.

Example 8: Find (α−β)2(\alpha - \beta)^2

If α,β\alpha, \beta are zeroes of x2−6x+8x^2 - 6x + 8, find (α−β)2(\alpha - \beta)^2.

Solution:

  1. Sum =6= 6, Product =8= 8.
  2. (α−β)2=(α+β)2−4αβ=62−4(8)=36−32=4(\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta = 6^2 - 4(8) = 36 - 32 = 4.

Final Answer: (α−β)2=4(\alpha - \beta)^2 = 4.

Takeaway: (α−β)2=(α+β)2−4αβ(\alpha-\beta)^2 = (\alpha+\beta)^2 - 4\alpha\beta.

Example 9: Sum of reciprocals squared

If α,β\alpha, \beta are zeroes of x2−5x+6x^2 - 5x + 6, find 1α2+1β2\dfrac{1}{\alpha^2} + \dfrac{1}{\beta^2}.

Solution:

  1. Sum =5= 5, Product =6= 6.
  2. 1α2+1β2=α2+β2(αβ)2\dfrac{1}{\alpha^2} + \dfrac{1}{\beta^2} = \dfrac{\alpha^2 + \beta^2}{(\alpha\beta)^2}.
  3. α2+β2=25−12=13\alpha^2 + \beta^2 = 25 - 12 = 13; (αβ)2=36(\alpha\beta)^2 = 36.
  4. So the value =1336= \dfrac{13}{36}.

Final Answer: 1336\dfrac{13}{36}.

Takeaway: Build the expression from sum and product step by step.

Example 10: Find a parameter using sum of zeroes

If the sum of the zeroes of 3x2+(2k+1)x−53x^2 + (2k+1)x - 5 is 1, find kk.

Solution:

  1. Sum of zeroes =−ba=−2k+13= -\dfrac{b}{a} = -\dfrac{2k+1}{3}.
  2. Set equal to 1: −2k+13=1⇒2k+1=−3-\dfrac{2k+1}{3} = 1 \Rightarrow 2k + 1 = -3.
  3. 2k=−4⇒k=−22k = -4 \Rightarrow k = -2.

Final Answer: k=−2k = -2.

Takeaway: Set the formula for the sum equal to the given value and solve for the unknown.