The zeroes of a quadratic polynomial are hidden in its coefficients: their sum is $-\tfrac ba$ and their product is $\tfrac ca$. Read them off, or run the idea backwards to build a quadratic with any zeroes you like.
Here is a puzzle to try on a friend. "I am thinking of two numbers. Their sum is $9$ and their product is $20$. What are they?" A little trial and error gives $4$ and $5$. Now look at the quadratic polynomial $x^2-9x+20$. It factorises as $(x-4)(x-5)$, so its zeroes are $4$ and $5$ โ the very numbers of the puzzle. The sum $9$ and the product $20$ were sitting in the coefficients all along.
This is no accident. In this section we prove that for every quadratic polynomial $ax^2+bx+c$ the sum of the zeroes is $-\tfrac ba$ and their product is $\tfrac ca$. That gives us three new skills: checking zeroes quickly, building a quadratic polynomial whose zeroes have a given sum and product, and working out expressions such as $\alpha^2+\beta^2$ without finding the zeroes at all. These skills come back in Chapter 4 (quadratic equations) and in many board and entrance-exam questions.
Finding zeroes by splitting the middle term
In Class 9 you factorised quadratics such as $x^2+7x+10$ by splitting the middle term: find two numbers whose sum is the coefficient of $x$ and whose product is (coefficient of $x^2$) ร (constant term). Once a quadratic is factorised, its zeroes come straight out, because a product is $0$ exactly when one of its factors is $0$.
โ๏ธ Example 1 โ Zeroes, then a pattern
Find the zeroes of $p(x)=x^2+7x+10$. Compare their sum and their product with the coefficients of $p(x)$.
๐ก Show solutionโถ
Split We need two numbers with sum $7$ and product $1\times10=10$: they are $2$ and $5$. So $x^2+7x+10=x^2+2x+5x+10=x(x+2)+5(x+2)=(x+2)(x+5)$.
Zeroes $(x+2)(x+5)=0$ when $x=-2$ or $x=-5$. The zeroes are $-2$ and $-5$.
Compare Here $a=1$, $b=7$, $c=10$. Sum of zeroes $=-2+(-5)=-7=-\dfrac71=-\dfrac ba$. Product of zeroes $=(-2)(-5)=10=\dfrac{10}{1}=\dfrac ca$.
โ๏ธ Example 2 โ When the leading coefficient is not 1
Find the zeroes of $6x^2-7x-3$ and verify the same pattern.
๐ก Show solutionโถ
Split We need two numbers with sum $-7$ and product $6\times(-3)=-18$: they are $-9$ and $2$. So
$$6x^2-7x-3=6x^2-9x+2x-3=3x(2x-3)+1(2x-3)=(3x+1)(2x-3).$$
Zeroes $3x+1=0$ gives $x=-\tfrac13$; $\;2x-3=0$ gives $x=\tfrac32$. The zeroes are $\tfrac32$ and $-\tfrac13$.
Sum $\tfrac32-\tfrac13=\tfrac96-\tfrac26=\tfrac76$, and $-\dfrac ba=-\dfrac{-7}{6}=\dfrac76$. โ
Product $\tfrac32\times\left(-\tfrac13\right)=-\tfrac12$, and $\dfrac ca=\dfrac{-3}{6}=-\dfrac12$. โ
The relationship between zeroes and coefficients
The pattern of Examples 1 and 2 always holds. To see why, we use the Factor Theorem from Class 9: if $p(\alpha)=0$, then $x-\alpha$ is a factor of $p(x)$.
๐ Theorem 2.2 โ Zeroes and coefficients of a quadratic polynomial
Let $p(x)=ax^2+bx+c$ with $a\ne0$, and let $\alpha$ and $\beta$ be its zeroes. Then
$$\alpha+\beta=-\frac ba=-\frac{\text{coefficient of }x}{\text{coefficient of }x^2},$$
$$\alpha\beta=\frac ca=\frac{\text{constant term}}{\text{coefficient of }x^2}.$$
(If the two zeroes are equal, take $\beta=\alpha$. If $p(x)$ has no real zeroes, the theorem says nothing.)
Why it is true. Since $\alpha$ and $\beta$ are zeroes, $x-\alpha$ and $x-\beta$ are factors of $p(x)$. A quadratic has degree $2$, so the only other factor can be a constant $k$: $\;p(x)=k(x-\alpha)(x-\beta)$. Multiplying out,
$$k(x-\alpha)(x-\beta)=kx^2-k(\alpha+\beta)x+k\alpha\beta .$$
Compare this with $ax^2+bx+c$, term by term. The $x^2$ terms give $k=a$. The $x$ terms give $b=-a(\alpha+\beta)$, so $\alpha+\beta=-\tfrac ba$. The constant terms give $c=a\alpha\beta$, so $\alpha\beta=\tfrac ca$.
โ ๏ธ Common mistakes
1. Losing the minus sign. The sum is $-\tfrac ba$, not $\tfrac ba$. For $x^2-5x+6$ the sum is $-\tfrac{-5}{1}=5$.
2. Forgetting to divide by $a$. For $2x^2-8x+6$ the product is $\tfrac62=3$, not $6$.
3. Reading coefficients from a jumbled polynomial. Rewrite $3-2x+x^2$ as $x^2-2x+3$ first: then $a=1$, $b=-2$, $c=3$ โ not $a=3$.
4. A missing term has coefficient $0$. In $t^2-15$, $b=0$ (so the sum is $0$); in $4u^2+8u$, $c=0$ (so the product is $0$).
Interactive 1 โ Move the Zeroes, Watch the Coefficients
Place the zeroes $\alpha$ and $\beta$ (red) and choose $a$. The graph is $y=a(x-\alpha)(x-\beta)$, and the readouts show its coefficients. The purple line sits halfway between the zeroes, at $x=\tfrac{\alpha+\beta}2$. The green dot is where the graph cuts the $y$-axis, at height $c=a\alpha\beta$. Press Play and watch: $\alpha+\beta$ always equals $-\tfrac ba$, and $\alpha\beta$ always equals $\tfrac ca$.
โ231
$p(x)=ax^2+bx+c$โ
$\alpha+\beta$ and $-\tfrac ba$โ
$\alpha\beta$ and $\tfrac ca$โ
Building a quadratic from the sum and product of its zeroes
Now run the idea backwards. Suppose we want a quadratic polynomial whose zeroes have sum $S$ and product $P$. The proof above shows that any such polynomial is $k(x-\alpha)(x-\beta)=k\left[x^2-(\alpha+\beta)x+\alpha\beta\right]$.
๐ Theorem 2.3 โ A quadratic with given sum and product of zeroes
If the zeroes $\alpha$, $\beta$ of a quadratic polynomial have sum $\alpha+\beta=S$ and product $\alpha\beta=P$, then the polynomial is
$$k\left[x^2-Sx+P\right]$$
for some real number $k\ne0$.
In words: $\;x^2-(\text{sum of zeroes})\,x+(\text{product of zeroes})$, multiplied by any non-zero constant.
The constant $k$ does not change the zeroes; it only stretches the graph. We usually choose $k$ to clear fractions, so that every coefficient is a whole number.
โ๏ธ Example 3 โ Building quadratic polynomials
Find a quadratic polynomial (a) whose zeroes have sum $\tfrac14$ and product $-1$; (b) whose zeroes have sum $\sqrt2$ and product $\tfrac13$; (c) whose zeroes are $2+\sqrt3$ and $2-\sqrt3$.
๐ก Show solutionโถ
(a) $x^2-\tfrac14x-1$. To clear the fraction take $k=4$: $\;4x^2-x-4$. Check: sum $=-\tfrac{-1}{4}=\tfrac14$ โ, product $=\tfrac{-4}{4}=-1$ โ.
(b) $x^2-\sqrt2\,x+\tfrac13$; take $k=3$: $\;3x^2-3\sqrt2\,x+1$. Check: sum $=\tfrac{3\sqrt2}{3}=\sqrt2$ โ, product $=\tfrac13$ โ. (Coefficients may be irrational โ that is allowed.)
(c) Sum and product $S=(2+\sqrt3)+(2-\sqrt3)=4$. $\;P=(2+\sqrt3)(2-\sqrt3)=2^2-(\sqrt3)^2=4-3=1$.
(c) Answer $x^2-4x+1$. The surds cancel, so the coefficients are whole numbers.
Note Each answer is "a" polynomial, not "the" polynomial: $8x^2-2x-8$ or $-4x^2+x+4$ would also do in (a), since $k$ can be any non-zero number.
Interactive 2 โ Two Numbers from Their Sum and Product
Two positive numbers $\alpha$ and $\beta$ with sum $S$ and product $P$ are the sides of a rectangle with half-perimeter $S$ and area $P$ (left). They are also the zeroes of $x^2-Sx+P$ (right). Press Play to increase the area $P$: the rectangle gets squarer, and the zeroes move together. When $P$ passes $\tfrac{S^2}4$ no such rectangle exists โ and the parabola lifts off the $x$-axis.
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polynomialโ
zeroes $\alpha$, $\beta$โ
check $\alpha+\beta$, $\alpha\beta$โ
Using the relationships without finding the zeroes
Many questions give you a quadratic and ask for an expression in its zeroes, such as $\alpha^2+\beta^2$. You could find $\alpha$ and $\beta$ first, but they are often ugly surds. It is much faster to rewrite the expression using only $\alpha+\beta$ and $\alpha\beta$, which you can read straight from the coefficients.
๐งญ Problem-solving strategy โ expressions in $\alpha$ and $\beta$
Write the polynomial in the form $ax^2+bx+c$ and note $\alpha+\beta=-\tfrac ba$ and $\alpha\beta=\tfrac ca$.
Rewrite the expression using these identities (from $(\alpha+\beta)^2=\alpha^2+2\alpha\beta+\beta^2$):
$$\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta$$
$$(\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta$$
$$\frac1\alpha+\frac1\beta=\frac{\alpha+\beta}{\alpha\beta}$$
Substitute the two values and simplify.
โ๏ธ Example 4 โ Values without the zeroes
Let $\alpha$ and $\beta$ be the zeroes of $p(x)=2x^2-6x+3$. Without finding $\alpha$ and $\beta$, find (a) $\dfrac1\alpha+\dfrac1\beta$ (b) $\alpha^2+\beta^2$ (c) the difference between the larger and the smaller zero.
๐ก Show solutionโถ
Read off $a=2$, $b=-6$, $c=3$: $\;\alpha+\beta=-\tfrac{-6}{2}=3$ and $\alpha\beta=\tfrac32$.
(c) $(\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta=9-6=3$, so the difference between the zeroes is $\sqrt3$.
Check In fact the zeroes are $\tfrac{3+\sqrt3}{2}\approx2.366$ and $\tfrac{3-\sqrt3}{2}\approx0.634$. Their difference is $1.732\approx\sqrt3$ โ, and $2.366^2+0.634^2\approx5.598+0.402=6$ โ. Imagine doing (b) with those surds โ the identities are much quicker.
โ๏ธ Example 5 โ Finding an unknown coefficient
(a) One zero of $x^2-6x+k$ is twice the other. Find $k$ and both zeroes.
(b) The zeroes of $3x^2-10x+k$ are reciprocals of each other (their product is $1$). Find $k$ and the zeroes.
๐ก Show solutionโถ
(a) Set up Call the zeroes $\alpha$ and $2\alpha$. Sum: $\alpha+2\alpha=-\tfrac{-6}{1}=6$, so $3\alpha=6$ and $\alpha=2$. The zeroes are $2$ and $4$.
(b) Reciprocals $\alpha$ and $\tfrac1\alpha$ have product $1$, so $\tfrac k3=1$ and $k=3$. Then $3x^2-10x+3=3x^2-9x-x+3=3x(x-3)-1(x-3)=(3x-1)(x-3)$: the zeroes are $\tfrac13$ and $3$, which are indeed reciprocals โ.
โ๏ธ Example 6 โ Looking ahead: three zeroes of a cubic
A cubic polynomial $ax^3+bx^2+cx+d$ has similar relations for its three zeroes $\alpha$, $\beta$, $\gamma$:
$$\alpha+\beta+\gamma=-\frac ba$$
$$\alpha\beta+\beta\gamma+\gamma\alpha=\frac ca$$
$$\alpha\beta\gamma=-\frac da$$
Verify that $3$, $-2$ and $\tfrac12$ are the zeroes of $2x^3-3x^2-11x+6$, and check all three relations. (This is a preview only; board questions on these relations are about quadratics.)
๐ก Show solutionโถ
Zeroes $p(3)=54-27-33+6=0$; $\;p(-2)=-16-12+22+6=0$; $\;p\!\left(\tfrac12\right)=\tfrac14-\tfrac34-\tfrac{11}2+6=0$. All three are zeroes, and a cubic has at most three, so these are all of them.
Sum $3+(-2)+\tfrac12=\tfrac32$, and $-\tfrac ba=-\tfrac{-3}{2}=\tfrac32$ โ.
Pairs $3(-2)+(-2)\left(\tfrac12\right)+\tfrac12(3)=-6-1+\tfrac32=-\tfrac{11}2$, and $\tfrac ca=-\tfrac{11}2$ โ.
Product $3\times(-2)\times\tfrac12=-3$, and $-\tfrac da=-\tfrac62=-3$ โ.
Exercises
I
Find the zeroes of each quadratic polynomial and verify the relationship between the zeroes and the coefficients.
7 We need two numbers with sum $2\sqrt2$ and product $-6$: $3\sqrt2$ and $-\sqrt2$. So $(x+3\sqrt2)(x-\sqrt2)$, zeroes $\sqrt2$, $-3\sqrt2$. Sum $-2\sqrt2=-\tfrac{2\sqrt2}{1}$ โ; product $-6$ โ.
Find a quadratic polynomial with whole-number coefficients (where possible) whose zeroes have the given sum and product, or the given zeroes.
1 sum $-3$, product $2$
2 sum $0$, product $-5$
3 sum $-\tfrac14$, product $-\tfrac38$
4 sum $2\sqrt3$, product $2$
5 zeroes $5$ and $-3$
6 zeroes $3+\sqrt2$ and $3-\sqrt2$
7 zeroes $-\tfrac23$ and $\tfrac12$
8 zeroes equal to the reciprocals of the zeroes of $x^2-5x+6$
Check: type a quadratic's coefficients and see its zeroes, their sum and their product
zeroesโ
sum $-\tfrac ba$โ
product $\tfrac ca$โ
๐ก Show solutionsโถ
1 $x^2-(-3)x+2=x^2+3x+2$ (zeroes $-1$, $-2$).
2 $x^2-0x-5=x^2-5$ (zeroes $\pm\sqrt5$).
3 $x^2+\tfrac14x-\tfrac38$; multiply by $8$: $\;8x^2+2x-3$ (zeroes $\tfrac12$, $-\tfrac34$).
4 $x^2-2\sqrt3\,x+2$ (zeroes $\sqrt3+1$ and $\sqrt3-1$: their sum is $2\sqrt3$ and product $3-1=2$).
5 $S=2$, $P=-15$: $\;x^2-2x-15$.
6 $S=6$, $P=9-2=7$: $\;x^2-6x+7$.
7 $S=-\tfrac23+\tfrac12=-\tfrac16$, $P=-\tfrac13$: $\;x^2+\tfrac16x-\tfrac13$; multiply by $6$: $\;6x^2+x-2=(3x+2)(2x-1)$ โ.
8 $x^2-5x+6=(x-2)(x-3)$, so the new zeroes are $\tfrac12$ and $\tfrac13$: $S=\tfrac56$, $P=\tfrac16$. Polynomial $x^2-\tfrac56x+\tfrac16$, or $6x^2-5x+1$.
III
Let $\alpha$ and $\beta$ be the zeroes of $p(x)=3x^2-6x+2$. Without finding $\alpha$ and $\beta$, find:
8 Sum $2\alpha+2\beta=4$, product $4\alpha\beta=\tfrac83$: $\;x^2-4x+\tfrac83$, or $3x^2-12x+8$.
Check The zeroes are $1\pm\tfrac1{\sqrt3}\approx1.577$ and $0.423$; for example $1.577^2+0.423^2\approx2.667=\tfrac83$ โ.
IV
Unknown coefficients and applications.
1 The sum of the zeroes of $kx^2+6x+4k$ is equal to their product. Find $k$.
2 One zero of $x^2-9x+k$ is twice the other. Find $k$.
3 The zeroes of $5x^2+13x+k$ are reciprocals of each other. Find $k$.
4 One zero of $2x^2+kx-6$ is $2$. Find $k$ and the other zero.
5 The zeroes of $x^2+px+12$ are positive whole numbers that differ by $1$. Find $p$.
6 The sum of the squares of the zeroes of $x^2-8x+k$ is $40$. Find $k$.
7Vegetable plot. A rectangular plot has perimeter $34$ m and area $60$ mยฒ. Write a quadratic polynomial whose zeroes are the length and breadth of the plot, and find them.
8Cricket ball. A ball hit upwards has height $h(t)=-5t^2+20t+25$ metres after $t$ seconds. Without solving, find the sum and product of the zeroes of $h(t)$. One zero is $-1$: find the other and say what it means.
๐ก Show solutionsโถ
1 Sum $-\tfrac6k$, product $\tfrac{4k}{k}=4$. So $-\tfrac6k=4$, $k=-\tfrac32$. (Check: $-\tfrac32x^2+6x-6=-\tfrac32(x-2)^2$ has zeroes $2$, $2$: sum $4$ = product $4$ โ.)
2 Zeroes $\alpha$, $2\alpha$: $3\alpha=9$, $\alpha=3$; zeroes $3$ and $6$, so $k=18$.
3 Product $=\tfrac k5=1$, so $k=5$.
4 $p(2)=8+2k-6=0\Rightarrow k=-1$. Product of zeroes $=\tfrac{-6}{2}=-3$, so the other zero is $\tfrac{-3}{2}=-\tfrac32$. (Check the sum: $2-\tfrac32=\tfrac12=-\tfrac{-1}{2}$ โ.)
5 Product $12$ with whole numbers differing by $1$: $3$ and $4$. Sum $7=-p$, so $p=-7$.
6 $\alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta=64-2k=40\Rightarrow k=12$. (Zeroes $2$ and $6$: $4+36=40$ โ.)
7 Length + breadth $=\tfrac{34}{2}=17$ m and length ร breadth $=60$ mยฒ. Polynomial $x^2-17x+60=(x-5)(x-12)$: the plot is $12$ m long and $5$ m wide.
8 Sum $=-\tfrac{20}{-5}=4$, product $=\tfrac{25}{-5}=-5$. If one zero is $-1$, the other is $4-(-1)=5$ (check: $-1\times5=-5$ โ). The ball hits the ground $5$ s after it is hit; $t=-1$ has no meaning here.
How this is examined: board papers ask you to find the zeroes and verify the relations, to build a quadratic from its zeroes, and to evaluate expressions such as $\alpha^2+\beta^2$ or $\tfrac1\alpha+\tfrac1\beta$. JEE Main uses the same identities in quick numerical questions. Write $\alpha+\beta=-\tfrac ba$ and $\alpha\beta=\tfrac ca$ as your first line โ examiners give marks for it.
E1.Board ยท 1 mark If $\alpha$ and $\beta$ are the zeroes of $x^2-5x+k$ and $\alpha-\beta=1$, then $k$ equals (A) $4$ (B) $6$ (C) $8$ (D) $-6$
๐ก Show solutionโถ
Idea $\alpha+\beta=5$ and $\alpha-\beta=1$. Adding, $2\alpha=6$, so $\alpha=3$ and $\beta=2$. Then $k=\alpha\beta=6$.
Trap (D) comes from taking the product as $-\tfrac ca$. The minus sign belongs to the sum, not the product.
Answer (B) $6$. Check: $x^2-5x+6=(x-2)(x-3)$ โ.
E2.Board ยท 1 mark A quadratic polynomial whose zeroes are $-3$ and $4$ is (A) $x^2-x+12$ (B) $x^2+x-12$ (C) $\tfrac{x^2}2-\tfrac x2-6$ (D) $2x^2+2x-24$
๐ก Show solutionโถ
Idea Sum $=1$, product $=-12$, so the polynomials are $k(x^2-x-12)$. With $k=\tfrac12$ this is $\tfrac{x^2}2-\tfrac x2-6$.
Traps (B) and (D) have the wrong sign of the middle term (they have zeroes $3$ and $-4$); (A) has the wrong sign of the product. Do not reject (C) just because it has fractions โ any non-zero $k$ is allowed.
Answer (C).
E3.Board ยท 1 markAssertion (A): A quadratic polynomial whose zeroes are $\tfrac12$ and $-2$ is $2x^2+3x-2$. Reason (R): The sum of the zeroes of $ax^2+bx+c$ is $\tfrac ba$. (A) Both A and R are true and R is the correct explanation of A. (B) Both A and R are true but R is not the correct explanation of A. (C) A is true but R is false. (D) A is false but R is true.
๐ก Show solutionโถ
Check A Sum $=\tfrac12-2=-\tfrac32$, product $=-1$: $\;x^2+\tfrac32x-1$, and $\times2$ gives $2x^2+3x-2$. True. (Also $(2x-1)(x+2)=2x^2+3x-2$ โ.)
Check R The sum is $-\tfrac ba$, not $\tfrac ba$. False.
Answer (C).
E4.Board ยท 2 marks Find the zeroes of $5x^2-4-8x$ and verify the relationship between the zeroes and the coefficients.
๐ก Show solutionโถ
Zeroes $5x^2-8x-4=5x^2-10x+2x-4=5x(x-2)+2(x-2)=(5x+2)(x-2)$. Zeroes $2$ and $-\tfrac25$. (1 mark)
E5.Board ยท 3 marks If $\alpha$ and $\beta$ are the zeroes of $4x^2-5x-1$, find a quadratic polynomial whose zeroes are $\dfrac1\alpha$ and $\dfrac1\beta$.
New sum and product $\dfrac1\alpha+\dfrac1\beta=\dfrac{\alpha+\beta}{\alpha\beta}=\dfrac{5/4}{-1/4}=-5$; $\;\dfrac1\alpha\cdot\dfrac1\beta=\dfrac1{\alpha\beta}=-4$. (1 mark)
Answer $x^2-(-5)x+(-4)=x^2+5x-4$ (or any $k(x^2+5x-4)$, $k\ne0$). (1 mark)
E6.JEE Main If $\alpha$ and $\beta$ are the zeroes of $x^2-6x+4$, then the value of $\alpha^3+\beta^3$ is ______. (numerical answer)
๐ก Show solutionโถ
Idea $\alpha+\beta=6$, $\alpha\beta=4$, and $\alpha^3+\beta^3=(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta)$.
Compute $6^3-3\times4\times6=216-72=144$.
Answer $144$. (Check with the zeroes $3\pm\sqrt5$: $(3+\sqrt5)^3+(3-\sqrt5)^3=2(27+3\times3\times5)=144$ โ.)
๐ฌ ResearchโTeaching NexusMini project ยท optional, about 40 minutes
Research question. For rectangles you find at home and at school, how close is the area to the largest area possible for the same perimeter?
Why it matters. Packaging designers, architects and farmers fencing a field all face the same question: with a fixed length of material (the perimeter), how much space (the area) can you enclose? The sides $\alpha$, $\beta$ of a rectangle with half-perimeter $S$ and area $P$ are the zeroes of $x^2-Sx+P$, and real zeroes exist only when $P\le\tfrac{S^2}4$ โ so the square is always the best rectangle.
What to do.
Measure the length and breadth (in cm) of $10$ rectangular objects: an exercise book, a phone, an A4 sheet, a door, a window pane, a whiteboard, a tileโฆ
For each one, work out $S=$ length $+$ breadth and $P=$ length $\times$ breadth, and write the polynomial $x^2-Sx+P$.
Check that the length and breadth really are the zeroes: substitute them into the polynomial (a spreadsheet makes this quick; expect small rounding errors).
Work out the "squareness" ratio $\dfrac{P}{S^2/4}=\dfrac{4P}{S^2}$ for each object. It is $1$ for a square and smaller for long thin rectangles.
Rank your objects by this ratio. Which shapes are designed to be close to square, and why might others be long and thin?
Tools: Tape measure or ruler, calculator or spreadsheet
Hand in: A table with length, breadth, $S$, $P$, the polynomial and the ratio $\tfrac{4P}{S^2}$ for each object, plus a short paragraph explaining why no rectangle can have a ratio bigger than $1$.
Going further. An A4 sheet measures about $21.0$ cm ร $29.7$ cm. Find its ratio, and compare it with that of a phone screen. Use Interactive 2 to find the area of the square with the same perimeter as the A4 sheet.
Teacher/parent note: since $S^2-4P=(\text{length}-\text{breadth})^2\ge0$, the ratio $\tfrac{4P}{S^2}$ is at most $1$, with equality only for a square. An A4 sheet gives about $0.971$; a typical phone (about $7$ cm ร $15$ cm) gives about $0.87$.
๐งช Self-check quiz
0/5 correct
Q1. The sum of the zeroes of $2x^2-8x+5$ is
Q2. The product of the zeroes of $3x^2+7x-6$ is
Q3. A quadratic polynomial whose zeroes have sum $-2$ and product $-15$ is
Q4. If $2$ is a zero of $x^2+kx+8$, the other zero is
Q5. If $\alpha$, $\beta$ are the zeroes of $x^2-3x+1$, then $\alpha^2+\beta^2$ equals