Relationship Between Zeroes and Coefficients, Class 10

Cubic poly­no­mi­als#

Let $\alpha$, $\beta$ and $\gamma$ be the zeroes of $p(x) = ax^3 + bx^2 + cx + d$, where $a \ne 0$. Then

$$\displaystyle \begin{aligned} \alpha + \beta + \gamma &= -\frac{b}{a} \\ \alpha\beta + \beta\gamma + \gamma\alpha &= \frac{c}{a} \\ \alpha\beta\gamma &= -\frac{d}{a} \end{aligned}$$

Notice the signs alter­nate: minus, plus, minus.

Why the results are true#

Proof for a qua­dratic#

If $\alpha$ and $\beta$ are the zeroes of $ax^2 + bx + c$, then $(x - \alpha)$ and $(x - \beta)$ are its fac­tors, and since the lead­ing coef­fi­cient is $a$,

$$ax^2 + bx + c = a(x - \alpha)(x - \beta)$$

Expand the right-hand side:

$$a(x - \alpha)(x - \beta) = ax^2 - a(\alpha + \beta)x + a\alpha\beta$$

Two poly­no­mi­als are equal for all $x$ only when their cor­re­spond­ing coef­fi­cients are equal. Com­par­ing the coef­fi­cients of $x$ and the con­stant terms:

$$\displaystyle \begin{aligned} b &= -a(\alpha + \beta) &&\Rightarrow\ \alpha + \beta = -\frac{b}{a} \\ c &= a\alpha\beta &&\Rightarrow\ \alpha\beta = \frac{c}{a} \end{aligned}$$

Proof for a cubic#

In the same way, $ax^3 + bx^2 + cx + d = a(x - \alpha)(x - \beta)(x - \gamma)$. Expand­ing,

$$a(x - \alpha)(x - \beta)(x - \gamma) = ax^3 - a(\alpha + \beta + \gamma)x^2 + a(\alpha\beta + \beta\gamma + \gamma\alpha)x - a\alpha\beta\gamma$$

Com­par­ing coef­fi­cients gives $b = -a(\alpha + \beta + \gamma)$, $c = a(\alpha\beta + \beta\gamma + \gamma\alpha)$ and $d = -a\alpha\beta\gamma$, which rearrange to the three results above.

Form­ing a qua­dratic from its zeroes#

The for­mula#

If a qua­dratic has zeroes whose sum is $S$ and prod­uct is $P$, then it is

$$k\left[x^2 - Sx + P\right], \quad k \ne 0$$

This fol­lows from $(x - \alpha)(x - \beta) = x^2 - (\alpha + \beta)x + \alpha\beta$. Any non-zero con­stant $k$ gives a qua­dratic with the same zeroes; choose $k$ to clear frac­tions.

Method#

1. Write down (or work out) $S = \alpha + \beta$ and $P = \alpha\beta$.
2. Write $x^2 - Sx + P$. Take care with signs: if $S$ is neg­a­tive, $-Sx$ becomes $+|S|x$.
3. Mul­ti­ply by the LCM of the denom­i­na­tors (this is your $k$) to get inte­ger coef­fi­cients.
4. Check: fac­torise or use $\displaystyle -\frac{b}{a}$ and $\displaystyle \frac{c}{a}$ to con­firm the sum and prod­uct.

Method: ver­i­fy­ing the rela­tion­ship#

1. Fac­torise the poly­no­mial (split the mid­dle term for a qua­dratic).
2. Set each fac­tor equal to $0$ to get the zeroes.
3. Work out the sum and the prod­uct of the zeroes you found.
4. Work out $\displaystyle -\frac{b}{a}$ and $\displaystyle \frac{c}{a}$ from the coef­fi­cients, keep­ing the signs of $b$ and $c$.
5. Show the two pairs of val­ues are equal.

Worked exam­ples#

Exam­ple 1: find and ver­ify#

Find the zeroes of $6x^2 - 5x - 6$ and ver­ify the rela­tion­ship with the coef­fi­cients.

We need two num­bers with prod­uct $6 \times (-6) = -36$ and sum $-5$: these are $-9$ and $4$.

$$\begin{aligned} 6x^2 - 5x - 6 &= 6x^2 - 9x + 4x - 6 \\ &= 3x(2x - 3) + 2(2x - 3) \\ &= (2x - 3)(3x + 2) \end{aligned}$$

So the zeroes are $\displaystyle \alpha = \frac{3}{2}$ and $\displaystyle \beta = -\frac{2}{3}$. Here $a = 6$, $b = -5$, $c = -6$.

$$\displaystyle \begin{aligned} \alpha + \beta &= \frac{3}{2} - \frac{2}{3} = \frac{9 - 4}{6} = \frac{5}{6}, &\quad -\frac{b}{a} &= -\frac{-5}{6} = \frac{5}{6} \\ \alpha\beta &= \frac{3}{2} \times \left(-\frac{2}{3}\right) = -1, &\quad \frac{c}{a} &= \frac{-6}{6} = -1 \end{aligned}$$

Both rela­tion­ships hold.

Exam­ple 2: a qua­dratic with no $x$ term#

Find the zeroes of $4x^2 - 9$ and ver­ify the rela­tion­ship.

$4x^2 - 9 = (2x - 3)(2x + 3)$, so the zeroes are $\displaystyle \frac{3}{2}$ and $\displaystyle -\frac{3}{2}$. Here $a = 4$, $b = 0$, $c = -9$.

$$\displaystyle \begin{aligned} \text{Sum} &= \frac{3}{2} + \left(-\frac{3}{2}\right) = 0 = -\frac{0}{4} \\ \text{Product} &= \frac{3}{2} \times \left(-\frac{3}{2}\right) = -\frac{9}{4} = \frac{c}{a} \end{aligned}$$

A miss­ing term sim­ply has coef­fi­cient $0$.

Exam­ple 3: form­ing qua­drat­ics#

Find a qua­dratic poly­no­mial whose zeroes have (a) sum $-4$ and prod­uct $3$, (b) sum $\displaystyle \frac{1}{4}$ and prod­uct $\displaystyle -\frac{1}{8}$.

(a) $x^2 - (-4)x + 3 = x^2 + 4x + 3$. Check: $x^2 + 4x + 3 = (x + 1)(x + 3)$, with zeroes $-1$ and $-3$; their sum is $-4$ and prod­uct is $3$.

(b) $\displaystyle x^2 - \frac{1}{4}x - \frac{1}{8}$. The LCM of $4$ and $8$ is $8$, so take $k = 8$:

$$\displaystyle 8\left[x^2 - \frac{1}{4}x - \frac{1}{8}\right] = 8x^2 - 2x - 1$$

Check: $8x^2 - 2x - 1 = (4x + 1)(2x - 1)$, with zeroes $\displaystyle -\frac{1}{4}$ and $\displaystyle \frac{1}{2}$. Sum $\displaystyle = \frac{1}{4}$ and prod­uct $\displaystyle = -\frac{1}{8}$, as required.

Exam­ple 4: irra­tional sum#

Find a qua­dratic poly­no­mial whose zeroes have sum $\sqrt{2}$ and prod­uct $-4$, and find its zeroes.

The poly­no­mial is $x^2 - \sqrt{2}x - 4$. To fac­torise, we need two num­bers with sum $-\sqrt{2}$ and prod­uct $-4$: these are $-2\sqrt{2}$ and $\sqrt{2}$.

$$\begin{aligned} x^2 - \sqrt{2}x - 4 &= x^2 - 2\sqrt{2}x + \sqrt{2}x - 4 \\ &= x(x - 2\sqrt{2}) + \sqrt{2}(x - 2\sqrt{2}) \\ &= (x - 2\sqrt{2})(x + \sqrt{2}) \end{aligned}$$

The zeroes are $2\sqrt{2}$ and $-\sqrt{2}$. Check: sum $= \sqrt{2}$ and prod­uct $= 2\sqrt{2} \times (-\sqrt{2}) = -4$.

Exam­ple 5: expres­sions in the zeroes#

If $\alpha$ and $\beta$ are the zeroes of $x^2 - 5x + 4$, find (a) $\displaystyle \frac{1}{\alpha} + \frac{1}{\beta}$, (b) $\alpha^2 + \beta^2$, (c) $(\alpha - \beta)^2$, with­out find­ing the zeroes.

Here $\displaystyle \alpha + \beta = -\frac{-5}{1} = 5$ and $\displaystyle \alpha\beta = \frac{4}{1} = 4$.

$$\displaystyle \begin{aligned} \text{(a)}\ \frac{1}{\alpha} + \frac{1}{\beta} &= \frac{\alpha + \beta}{\alpha\beta} = \frac{5}{4} \\ \text{(b)}\ \alpha^2 + \beta^2 &= (\alpha + \beta)^2 - 2\alpha\beta = 25 - 8 = 17 \\ \text{(c)}\ (\alpha - \beta)^2 &= (\alpha + \beta)^2 - 4\alpha\beta = 25 - 16 = 9 \end{aligned}$$

Check: $x^2 - 5x + 4 = (x - 1)(x - 4)$, so the zeroes are $1$ and $4$. Then $\displaystyle 1 + \frac{1}{4} = \frac{5}{4}$, $1 + 16 = 17$ and $(1 - 4)^2 = 9$.

Exam­ple 6: a cubic, ver­i­fied#

Show that $3$, $\displaystyle \frac{1}{2}$ and $-1$ are the zeroes of $p(x) = 2x^3 - 5x^2 - 4x + 3$, and ver­ify the three rela­tion­ships.

First, sub­sti­tute:

$$\displaystyle \begin{aligned} p(3) &= 2(27) - 5(9) - 4(3) + 3 = 54 - 45 - 12 + 3 = 0 \\ p\left(\tfrac{1}{2}\right) &= 2\left(\tfrac{1}{8}\right) - 5\left(\tfrac{1}{4}\right) - 2 + 3 = \tfrac{1}{4} - \tfrac{5}{4} + 1 = 0 \\ p(-1) &= 2(-1) - 5(1) + 4 + 3 = -2 - 5 + 4 + 3 = 0 \end{aligned}$$

So all three are zeroes; in fact $p(x) = (x - 3)(2x - 1)(x + 1)$. Now $a = 2$, $b = -5$, $c = -4$, $d = 3$. Take $\alpha = 3$, $\displaystyle \beta = \frac{1}{2}$, $\gamma = -1$.

$$\displaystyle \begin{aligned} \alpha + \beta + \gamma &= 3 + \tfrac{1}{2} - 1 = \tfrac{5}{2} = -\frac{-5}{2} = -\frac{b}{a} \\ \alpha\beta + \beta\gamma + \gamma\alpha &= \tfrac{3}{2} + \left(-\tfrac{1}{2}\right) + (-3) = -2 = \frac{-4}{2} = \frac{c}{a} \\ \alpha\beta\gamma &= 3 \times \tfrac{1}{2} \times (-1) = -\tfrac{3}{2} = -\frac{3}{2} = -\frac{d}{a} \end{aligned}$$

All three rela­tion­ships hold.

Com­mon mis­takes#

• Writ­ing the sum as $\displaystyle \frac{b}{a}$ instead of $\displaystyle -\frac{b}{a}$.
• Ignor­ing the sign of $b$ or $c$: for $6x^2 - 5x - 6$, $b = -5$ and $c = -6$, not $5$ and $6$.
• Form­ing the qua­dratic as $x^2 + Sx + P$. The mid­dle sign is minus: $x^2 - Sx + P$.
• Using the cubic prod­uct rule with­out the minus sign: $\displaystyle \alpha\beta\gamma = -\frac{d}{a}$.
• Writ­ing $\alpha^2 + \beta^2 = (\alpha + \beta)^2$; the term $-2\alpha\beta$ is needed.
• For­get­ting that a miss­ing term has coef­fi­cient $0$.
• Say­ing "the" qua­dratic when the answer is only fixed up to a non-zero con­stant $k$.

Try these#

1. Find the sum and prod­uct of the zeroes of $3x^2 + 7x - 6$. Answer: sum $\displaystyle -\frac{7}{3}$, prod­uct $-2$ (zeroes $\displaystyle \frac{2}{3}$ and $-3$).
2. Find a qua­dratic poly­no­mial whose zeroes have sum $\displaystyle -\frac{1}{2}$ and prod­uct $-3$. Answer: $2x^2 + x - 6$.
3. One zero of $x^2 + kx - 15$ is $3$. Find $k$ and the other zero. Answer: $k = 2$; the other zero is $-5$.
4. If $\alpha$ and $\beta$ are the zeroes of $x^2 - 7x + 10$, find $\alpha^2 + \beta^2$. Answer: $29$.
5. The zeroes of $x^3 - 6x^2 + 11x - 6$ are $1$, $2$ and $3$. Give the sum, the sum of prod­ucts two at a time, and the prod­uct. Answer: $6$, $11$ and $6$.

Key terms#

Zero

A value of $x$ for which the poly­no­mial equals $0$.

Coef­fi­cient

The num­ber mul­ti­ply­ing a power of $x$ in a term.

Lead­ing coef­fi­cient

The coef­fi­cient of the high­est power, writ­ten $a$; it is never $0$.

Con­stant term

The term with­out $x$ ($c$ in a qua­dratic, $d$ in a cubic).

Sum of zeroes

$\displaystyle -\frac{b}{a}$ for both a qua­dratic and a cubic.

Prod­uct of zeroes

$\displaystyle \frac{c}{a}$ for a qua­dratic and $\displaystyle -\frac{d}{a}$ for a cubic.

Sum of prod­ucts two at a time

$\displaystyle \alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a}$ for a cubic.

Fac­tor

If $\alpha$ is a zero, then $(x - \alpha)$ divides the poly­no­mial exactly.

Com­mon ques­tions#

Why is there a $k$ when form­ing a qua­dratic?#

Mul­ti­ply­ing a poly­no­mial by any non-zero num­ber does not change where it equals $0$, so infi­nitely many qua­drat­ics share the same zeroes. They dif­fer only by the fac­tor $k$.

Do the rela­tion­ships hold if the zeroes are equal?#

Yes. For $(x - 3)^2 = x^2 - 6x + 9$, the zeroes are $3$ and $3$: sum $\displaystyle 6 = -\frac{-6}{1}$ and prod­uct $\displaystyle 9 = \frac{9}{1}$.

Can I use the rela­tion­ships instead of fac­toris­ing?#

For sums, prod­ucts and expres­sions such as $\alpha^2 + \beta^2$, yes. To find the actual zeroes you still need to fac­torise or solve.

How do I remem­ber the cubic signs?#

Divide the coef­fi­cients by $a$ and read them in order, attach­ing the signs minus, plus, minus: $\displaystyle -\frac{b}{a}$, $\displaystyle +\frac{c}{a}$, $\displaystyle -\frac{d}{a}$.

Is the order of $\alpha$ and $\beta$ impor­tant?#

No. Sums and prod­ucts do not depend on the order, so either zero can be called $\alpha$.

Ref­er­ences#

1. National Coun­cil of Edu­ca­tional Research and Train­ing. Math­e­mat­ics: Text­book for Class X. NCERT, New Delhi.
2. Aggar­wal, R. S. Sec­ondary School Math­e­mat­ics for Class 10. Bharati Bhawan.
3. Hall, H. S. and Knight, S. R. Higher Alge­bra. Macmil­lan.