38 MATHEMATICS
4
4.1 Introduction
In Chapter 2, you have studied different types of polynomials. One type was the
quadratic polynomial of the form ax
2
+ bx + c, a 0. When we equate this polynomial
to zero, we get a quadratic equation. Quadratic equations come up when we deal with
many real-life situations. For instance, suppose a
charity trust decides to build a prayer hall having
a carpet area of 300 square metres with its length
one metre more than twice its breadth. What
should be the length and breadth of the hall?
Suppose the breadth of the hall is x metres. Then,
its length should be (2x + 1) metres. We can depict
this information pictorially as shown in Fig. 4.1.
Now, area of the hall = (2x + 1). x m
2
= (2x
2
+ x) m
2
So, 2x
2
+ x = 300 (Given)
Therefore, 2x
2
+ x – 300 = 0
So, the breadth of the hall should satisfy the equation 2x
2
+ x – 300 = 0 which is a
quadratic equation.
Many people believe that Babylonians were the first to solve quadratic equations.
For instance, they knew how to find two positive numbers with a given positive sum
and a given positive product, and this problem is equivalent to solving a quadratic
equation of the form x
2
– px + q = 0. Greek mathematician Euclid developed a
geometrical approach for finding out lengths which, in our present day terminology,
are solutions of quadratic equations. Solving of quadratic equations, in general form, is
often credited to ancient Indian mathematicians. In fact, Brahmagupta (C.E.598–665)
gave an explicit formula to solve a quadratic equation of the form ax
2
+ bx = c. Later,
QUADRATIC EQUATIONS
Fig. 4.1
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QUADRATIC EQUATIONS 39
Sridharacharya (C.E. 1025) derived a formula, now known as the quadratic formula,
(as quoted by Bhaskara II) for solving a quadratic equation by the method of completing
the square. An Arab mathematician Al-Khwarizmi (about C.E. 800) also studied
quadratic equations of different types. Abraham bar Hiyya Ha-Nasi, in his book
‘Liber embadorum’ published in Europe in C.E. 1145 gave complete solutions of
different quadratic equations.
In this chapter, you will study quadratic equations, and various ways of finding
their roots. You will also see some applications of quadratic equations in daily life
situations.
4.2 Quadratic Equations
A quadratic equation in the variable x is an equation of the form ax
2
+ bx + c = 0, where
a, b, c are real numbers, a 0. For example, 2x
2
+ x – 300 = 0 is a quadratic equation.
Similarly, 2x
2
– 3x + 1 = 0, 4x – 3x
2
+ 2 = 0 and 1 – x
2
+ 300 = 0 are also quadratic
equations.
In fact, any equation of the form p(x) = 0, where p(x) is a polynomial of degree
2, is a quadratic equation. But when we write the terms of p(x) in descending order of
their degrees, then we get the standard form of the equation. That is, ax
2
+ bx + c = 0,
a 0 is called the standard form of a quadratic equation.
Quadratic equations arise in several situations in the world around us and in
different fields of mathematics. Let us consider a few examples.
Example 1 : Represent the following situations mathematically:
(i) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and
the product of the number of marbles they now have is 124. W
e would like to find
out how many marbles they had to start with.
(ii) A cottage industry produces a certain number of toys in a day. The cost of
production of each toy (in rupees) was found to be 55 minus the number of toys
produced in a day. On a particular day, the total cost of production was
` 750. We would like to find out the number of toys produced on that day.
Solution :
(i) Let the number of marbles John had be x.
Then the number of marbles Jivanti had = 45 – x (Why?).
The number of marbles left with John, when he lost 5 marbles = x – 5
The number of marbles left with Jivanti, when she lost 5 marbles = 45 – x – 5
= 40 – x
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40 MATHEMATICS
Therefore, their product = (x – 5) (40 – x)
=40x – x
2
– 200 + 5x
=– x
2
+ 45x – 200
So, – x
2
+ 45x – 200 = 124 (Given that product = 124)
i.e., – x
2
+ 45x – 324 = 0
i.e., x
2
– 45x + 324 = 0
Therefore, the number of marbles John had, satisfies the quadratic equation
x
2
– 45x + 324 = 0
which is the required representation of the problem mathematically.
(ii) Let the number of toys produced on that day be x.
Therefore, the cost of production (in rupees) of each toy that day = 55 – x
So, the total cost of production (in rupees) that day = x (55 – x)
Therefore, x (55 – x) = 750
i.e., 55x – x
2
= 750
i.e., – x
2
+ 55x – 750 = 0
i.e., x
2
– 55x + 750 = 0
Therefore, the number of toys produced that day satisfies the quadratic equation
x
2
– 55x + 750 = 0
which is the required representation of the problem mathematically.
Example 2 : Check whether the following are quadratic equations:
(i) (x – 2)
2
+ 1 = 2x – 3 (ii) x(x + 1) + 8 = (x + 2) (x – 2)
(iii) x (2x + 3) = x
2
+ 1 (iv) (x + 2)
3
= x
3
– 4
Solution :
(i) LHS = (x – 2)
2
+ 1 = x
2
– 4x + 4 + 1 = x
2
– 4x + 5
Therefore, (x – 2)
2
+
1 = 2x – 3 can be rewritten as
x
2
– 4x + 5 = 2x – 3
i.e., x
2
– 6x + 8 = 0
It is of the form ax
2
+ bx + c = 0.
Therefore, the given equation is a quadratic equation.
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QUADRATIC EQUATIONS 41
(ii) Since x(x + 1) + 8 = x
2
+ x + 8 and (x + 2)(x – 2) = x
2
– 4
Therefore, x
2
+ x + 8 = x
2
– 4
i.e., x + 12 = 0
It is not of the form ax
2
+ bx + c = 0.
Therefore, the given equation is not a quadratic equation.
(iii) Here, LHS = x (2x + 3) = 2x
2
+ 3x
So, x (2x + 3) = x
2
+ 1 can be rewritten as
2x
2
+ 3x = x
2
+ 1
Therefore, we get x
2
+ 3x – 1 = 0
It is of the form ax
2
+ bx + c = 0.
So, the given equation is a quadratic equation.
(iv) Here, LHS = (x + 2)
3
= x
3
+ 6x
2
+ 12x + 8
Therefore, (x + 2)
3
= x
3
– 4 can be rewritten as
x
3
+ 6x
2
+ 12x + 8 = x
3
– 4
i.e., 6x
2
+ 12x + 12 = 0 or, x
2
+ 2x + 2 = 0
It is of the form ax
2
+ bx + c = 0.
So, the given equation is a quadratic equation.
Remark : Be careful! In (ii) above, the given equation appears to be a quadratic
equation, but it is not a quadratic equation.
In (iv) above, the given equation appears to be a cubic equation (an equation of
degree 3) and not a quadratic equation. But it turns out to be a quadratic equation.
As
you can see, often we need to simplify the given equation before deciding whether it
is quadratic or not.
EXERCISE 4.1
1. Check whether the following are quadratic equations :
(i) (x + 1)
2
= 2(x – 3) (ii) x
2
– 2x = (–2) (3 – x)
(iii) (x – 2)(x + 1) = (x – 1)(x + 3) (iv) (x – 3)(2x +1) = x(x + 5)
(v) (2x – 1)(x – 3) = (x + 5)(x – 1) (vi) x
2
+ 3x + 1 = (x – 2)
2
(vii) (x + 2)
3
= 2x (x
2
– 1) (viii) x
3
– 4x
2
– x + 1 = (x – 2)
3
2. Represent the following situations in the form of quadratic equations :
(i) The area of a rectangular plot is 528 m
2
. The length of the plot (in metres) is one
more than twice its breadth. We need to find the length and breadth of the plot.
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42 MATHEMATICS
(ii) The product of two consecutive positive integers is 306. We need to find the
integers.
(iii) Rohan’s mother is 26 years older than him. The product of their ages (in years)
3 years from now will be 360. We would like to find Rohan’s present age.
(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been
8 km/h less, then it would have taken 3 hours more to cover the same distance. We
need to find the speed of the train.
4.3 Solution of a Quadratic Equation by Factorisation
Consider the quadratic equation 2x
2
– 3x + 1 = 0. If we replace x by 1 on the
LHS of this equation, we get (2 × 1
2
) – (3 × 1) + 1 = 0 = RHS of the equation.
We say that 1 is a root of the quadratic equation 2x
2
– 3x + 1 = 0. This also means that
1 is a zero of the quadratic polynomial 2x
2
– 3x + 1.
In general, a real number is called a root of the quadratic equation
ax
2
+ bx + c = 0, a 0 if a
2
+ b + c = 0. We also say that x =
is a solution of
the quadratic equation, or that
satisfies the quadratic equation. Note that the
zeroes of the quadratic polynomial ax
2
+ bx + c and the roots of the quadratic
equation ax
2
+ bx + c = 0 are the same.
You have observed, in Chapter 2, that a quadratic polynomial can have at most
two zeroes. So, any quadratic equation can have atmost two roots.
You have learnt in Class IX, how to factorise quadratic polynomials by splitting
their middle terms. We shall use this knowledge for finding the roots of a quadratic
equation. Let us see how.
Example 3 : Find the roots of the equation 2x
2
– 5x + 3 = 0, by factorisation.
Solution : Let us first split the middle term – 5x as –2x –3x [because (–2x) × (–3x) =
6x
2
= (2x
2
) × 3].
So, 2x
2
– 5x + 3 = 2x
2
– 2x – 3x + 3 = 2x (x – 1) –3(x – 1) = (2x – 3)(x – 1)
Now, 2x
2
– 5x + 3 = 0 can be rewritten as (2x – 3)(x – 1) = 0.
So, the values of x for which 2x
2
– 5x + 3 = 0 are the same for which (2x – 3)(x – 1) = 0,
i.e., either 2x – 3 = 0 or x – 1 = 0.
Now, 2x – 3 = 0 gives
3
2
x
and x – 1 = 0 gives x = 1.
So,
3
2
x
and x = 1 are the solutions of the equation.
In other words, 1 and
3
2
are the roots of the equation 2x
2
– 5x + 3 = 0.
Verify that these are the roots of the given equation.
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QUADRATIC EQUATIONS 43
Note that we have found the roots of 2x
2
– 5x + 3 = 0 by factorising
2x
2
– 5x + 3 into two linear factors and equating each factor to zero.
Example 4 : Find the roots of the quadratic equation 6x
2
– x – 2 = 0.
Solution : We have
6x
2
– x – 2 = 6x
2
+ 3x – 4x – 2
=3x (2x + 1) – 2 (2x + 1)
=(3x – 2)(2x + 1)
The roots of 6x
2
– x – 2 = 0 are the values of x for which (3x – 2)(2x + 1) = 0
Therefore, 3x – 2 = 0 or 2x + 1 = 0,
i.e., x =
2
3
or x =
1
2
Therefore, the roots of 6x
2
– x – 2 = 0 are
21
.
and –
32
We verify the roots, by checking that
21
and
32
satisfy 6x
2
– x – 2 = 0.
Example 5 : Find the roots of the quadratic equation
2
32620xx
.
Solution :
2
3262xx
=
2
3662xxx
=
33 2 23 2xx x
=
3232xx
So, the roots of the equation are the values of x for which
323 20xx
Now,
320x
for
2
3
x
.
So, this root is repeated twice, one for each repeated factor
32x
.
Therefore, the roots of
2
32620xx
are
2
3
,
2
3
.
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44 MATHEMATICS
Example 6 : Find the dimensions of the prayer hall discussed in Section 4.1.
Solution : In Section 4.1, we found that if the breadth of the hall is x m, then x
satisfies the equation 2x
2
+ x – 300 = 0. Applying the factorisation method, we write
this equation as
2x
2
– 24x + 25x – 300 = 0
2x (x – 12) + 25 (x – 12) = 0
i.e., (x – 12)(2x + 25) = 0
So, the roots of the given equation are x = 12 or x = – 12.5. Since x is the breadth
of the hall, it cannot be negative.
Thus, the breadth of the hall is 12 m. Its length = 2x + 1 = 25 m.
EXERCISE 4.2
1. Find the roots of the following quadratic equations by factorisation:
(i) x
2
– 3x – 10 = 0 (ii) 2x
2
+ x – 6 = 0
(iii)
2
27520xx
(iv) 2x
2
– x +
1
8
= 0
(v) 100 x
2
– 20x + 1 = 0
2. Solve the problems given in Example 1.
3. Find two numbers whose sum is 27 and product is 182.
4. Find two consecutive positive integers, sum of whose squares is 365.
5. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find
the other two sides.
6. A cottage industry produces a certain number of pottery articles in a day. It was observed
on a particular day that the cost of production of each article (in rupees) was 3 more than
twice the number of articles produced on that day. If the total cost of production on that
day was ` 90, find the number of articles produced and the cost of each article.
4.4 Nature of Roots
The equation ax
2
+ bx + c = 0 are given by
x =
2
–4
2
bbac
a
If b
2
– 4ac > 0, we get two distinct real roots
2
4
22
bac
b
aa
and
2
4
–
22
bac
b
aa
.
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QUADRATIC EQUATIONS 45
If b
2
– 4ac = 0, then x =
0
2
b
a
,
i.e., or –
22
bb
x
aa
So, the roots of the equation ax
2
+ bx + c = 0 are both
2
b
a
Therefore, we say that the quadratic equation ax
2
+ bx + c = 0 has two equal
real roots in this case.
If b
2
– 4ac < 0, then there is no real number whose square is b
2
– 4ac. Therefore,
there are no real roots for the given quadratic equation in this case.
Since b
2
– 4ac determines whether the quadratic equation ax
2
+ bx + c = 0 has
real roots or not, b
2
– 4ac is called the discriminant of this quadratic equation.
So, a quadratic equation ax
2
+ bx + c = 0 has
(i) two distinct real roots, if b
2
– 4ac > 0,
(ii) two equal real roots, if b
2
– 4ac = 0,
(iii)
no real roots, if b
2
– 4ac < 0.
Let us consider some examples.
Example 7: Find the discriminant of the quadratic equation 2x
2
– 4x + 3 = 0, and
hence find the nature of its roots.
Solution : The given equation is of the form ax
2
+ bx + c = 0, where a = 2, b = – 4 and
c = 3. Therefore, the discriminant
b
2
– 4ac = (– 4)
2
– (4 × 2 × 3) = 16 – 24 = – 8 < 0
So, the given equation has no real roots.
Example 8 : A pole has to be erected at a point on the boundary of a circular park of
diameter 13 metres in such a way that the differences of its distances from two
diametrically opposite fixed gates A and B on the boundary is 7 metres. Is it possible to
do so? If yes, at what distances from the two gates should the pole be erected?
Solution : Let us first draw the diagram
(see Fig. 4.2).
Let P be the required location of the
pole. Let the distance of the pole from the
gate B be x m, i.e., BP = x m. Now the
difference of the distances of the pole from
the two gates = AP
– BP (or, BP
– AP) =
7 m. Therefore, AP = (x + 7) m.
Fig. 4.2
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46 MATHEMATICS
Now, AB = 13m, and since AB is a diameter,
APB = 90° (Why?)
Therefore, AP
2
+ PB
2
=AB
2
(By Pythagoras theorem)
i.e., (x + 7)
2
+ x
2
=13
2
i.e., x
2
+ 14x + 49 + x
2
= 169
i.e., 2x
2
+ 14x – 120 = 0
So, the distance ‘x’ of the pole from gate B satisfies the equation
x
2
+ 7x – 60 = 0
So, it would be possible to place the pole if this equation has real roots. To see if this
is so or not, let us consider its discriminant. The discriminant is
b
2
– 4ac = 7
2
– 4 × 1 × (– 60) = 289 > 0.
So, the given quadratic equation has two real roots, and it is possible to erect the
pole on the boundary of the park.
Solving the quadratic equation x
2
+ 7x – 60 = 0, by the quadratic formula, we get
x =
7289
2
=
717
2
Therefore, x = 5 or – 12.
Since x is the distance between the pole and the gate B, it must be positive.
Therefore, x = – 12 will have to be ignored. So, x = 5.
Thus, the pole has to be erected on the boundary of the park at a distance of 5m
from the gate B and 12m from the gate A.
Example 9 : Find the discriminant of the equation 3x
2
– 2x +
1
3
= 0 and hence find the
nature of its roots. Find them, if they are real.
Solution : Here a = 3, b = – 2 and
1
3
c
.
Therefore, discriminant b
2
– 4ac = (– 2)
2
– 4 × 3 ×
1
3
= 4 – 4 = 0.
Hence, the given quadratic equation has two equal real roots.
The roots are
22 11
,,
,,
.
i.e., , i.e.,
22 66 33
bb
aa
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QUADRATIC EQUATIONS 47
EXERCISE 4.3
1. Find the nature of the roots of the following quadratic equations. If the real roots exist,
find them:
(i) 2x
2
– 3x + 5 = 0 (ii) 3x
2
– 4
3
x + 4 = 0
(iii) 2x
2
– 6x + 3 = 0
2. Find the values of k for each of the following quadratic equations, so that they have two
equal roots.
(i) 2x
2
+ kx + 3 = 0 (ii) kx (x – 2) + 6 = 0
3. Is it possible to design a rectangular mango grove whose length is twice its breadth,
and the area is 800 m
2
? If so, find its length and breadth.
4. Is the following situation possible? If so, determine their present ages.
The sum of the ages of two friends is 20 years. Four years ago, the product of their ages
in years was 48.
5. Is it possible to design a rectangular park of perimeter 80 m and area 400 m
2
? If so, find
its length and breadth.
4.5 Summary
In this chapter, you have studied the following points:
1. A quadratic equation in the variable x is of the form ax
2
+ bx + c = 0, where a, b, c are real
numbers and a 0.
2. A real number is said to be a root of the quadratic equation ax
2
+ bx + c = 0, if
a
2
+ b + c = 0. The zeroes of the quadratic polynomial ax
2
+ bx + c and the roots of the
quadratic equation ax
2
+ bx + c = 0 are the same.
3. If we can factorise ax
2
+ bx + c, a 0, into a product of two linear factors, then the roots
of the quadratic equation ax
2
+ bx + c = 0 can be found by equating each factor to zero.
4. Quadratic formula: The roots of a quadratic equation ax
2
+ bx + c = 0 are given by
2
4
,
2
bb ac
a
provided b
2
– 4ac 0.
5. A quadratic equation ax
2
+ bx + c = 0 has
(i) two distinct real roots, if b
2
– 4ac > 0,
(ii) two equal roots (i.e., coincident roots), if b
2
– 4ac = 0, and
(iii) no real roots, if b
2
– 4ac < 0.
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48 MATHEMATICS
NOTE
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