QUDRATIC EQUATIONS

1. QUADRATIC EQUATIONS
QUADRATIC
EQUATIONS

2. QUADRATIC EQUATIONS

3. QUADRATIC EQUATIONS
Theorem: A necessary and sufficient condition for the
quadratic equations a1x2+b1x+c1=0 and a2x2+b2x+c2=0 to
have a common root is c1a2−c2a1
2
= a1b2−a2b1 b1c2−b2c1
Necessity
Let  be a common root of the given equations then
a1
2
+b1 +c1= 0 1
a2
2
+b2 +c2= 0  2
Proof

4. QUADRATIC EQUATIONS
Solving (1) & (2)
b1 c1
b2 c2 a2
a1 b1
b2
By the method of cross multiplication,
2  1
⇒
2
b𝟏c𝟐−b𝟐c𝟏
=

c𝟐a𝟐−c𝟐a𝟏
=
1
a𝟏b𝟐−a𝟐b𝟏
⇒2 =
b𝟏c𝟐−b𝟐c𝟏
a𝟏b𝟐−a𝟐b𝟏
,  =
c𝟏a𝟐−c𝟐a𝟏
a𝟏b𝟐−a𝟐b𝟏

5. QUADRATIC EQUATIONS
⇒
c𝟏a𝟐−c𝟐a𝟏
a𝟏b𝟐−a𝟐b𝟏
2
=
b𝟏c𝟐−b𝟐c𝟏
a𝟏b𝟐−a𝟐b𝟏
⇒ c𝟏a𝟐−c𝟐a𝟏
𝟐 = a𝟏b𝟐−a𝟐b𝟏 b1 c2 − b2 c1
Sufficiency
Suppose that
Case (i)
a𝟏b𝟐−a𝟐b𝟏 b1 c2 − b2 c1 = c𝟏a𝟐−c𝟐a𝟏
𝟐 5
a1b2 – a2b1=0 then c1a2-c2a1=0 (from 5)

6. QUADRATIC EQUATIONS
i.e.,
a𝟏
a𝟐
=
b𝟏
b𝟐
=
c𝟏
c𝟐
⇒ a1x2+b1x+c1= 0, a2x2+b2x+c2 = 0 have the same
roots and hence these equations have a common
root.
⇒
a𝟏
a𝟐
=
b𝟏
b𝟐
and
a𝟏
a𝟐
=
c𝟏
c𝟐

7. QUADRATIC EQUATIONS
If a𝟏b𝟐−a𝟐b𝟏  0 then let  =
c𝟏a𝟐−c𝟐a𝟏
a𝟏b𝟐−a𝟐b𝟏
.
=
a𝟏(c𝟏a𝟐−c𝟐a𝟏)𝟐
+b𝟏 c𝟏a𝟐−c𝟐a𝟏 a𝟏b𝟐−a𝟐b𝟏 + c𝟏 a𝟏b𝟐−a𝟐b𝟏
𝟐
a𝟏b𝟐−a𝟐b𝟏
𝟐
Case (ii)
Now a12+b1+c1 = a1
c𝟏a𝟐−c𝟐a𝟏
a𝟏b𝟐−a𝟐b𝟏
𝟐
+b𝟏
c𝟏a𝟐−c𝟐a𝟏
a𝟏b𝟐−a𝟐b𝟏
+ c𝟏

8. QUADRATIC EQUATIONS
=
a𝟏(a𝟏b𝟐−a𝟐b𝟏)(b𝟏c𝟐−b𝟐c𝟏) + b𝟏 c𝟏a𝟐−c𝟐a𝟏 a𝟏b𝟐−a𝟐b𝟏 + c𝟏 a𝟏b𝟐−a𝟐b𝟏
𝟐
a𝟏b𝟐−a𝟐b𝟏
𝟐
=
a𝟏 (b𝟏c𝟐−b𝟐c𝟏) + b𝟏 c𝟏a𝟐−c𝟐a𝟏 + c𝟏 a𝟏b𝟐−a𝟐b𝟏
a𝟏b𝟐−a𝟐b𝟏
=
0
a𝟏b𝟐−a𝟐b𝟏
= 0
i.e., a12+b1+c1=0 similarly we can prove that
a22+b2+c2=0.
Thus, ‘’ is a common root of the given equations

9. QUADRATIC EQUATIONS
1) If x2+4ax+3=0 and 2x2+3ax-9=0 have a common root,
then find the values of ‘a’ and the common roots.
Solution
a1=1, b1=4a, c1=3,
c𝟏a𝟐−c𝟐a𝟏
𝟐= a𝟏b𝟐−a𝟐b𝟏 b1c2 − b2c1 .
⇒ 3 2 − −9 1 𝟐
a2=2, b2=3a, c2=-9
are substitute in the condition
Here
= 1 3a −2 4a 4a −9 − 3a 3

10. QUADRATIC EQUATIONS
i.e., 225 = (-5a)(-45a)
The common root =
c𝟏a𝟐−c𝟐a𝟏
a𝟏b𝟐−a𝟐b𝟏
i.e., a2=1 So, a=1
=
3 2 − −9 1
1 3a − 2 4a
=
15
−5a =
−3
a
If a=-1, the common root = 3,
If a=1, the common root = -3

11. QUADRATIC EQUATIONS

12. QUADRATIC EQUATIONS
EQUATIONS REDUCIBLE TO QUADRATIC EQUATIONS
Here is an explanation with some illustrations how to
solve some equations which are reducible to quadratic
equations by suitable substitutions.

13. QUADRATIC EQUATIONS
1) Solve 71+x+71-x-50 = 0
Solution
The given equation can be written as 7.7x +
7
7
𝐱 - 50 =0
i.e, 7. 7x 𝟐
- 50.7x+7 = 0
On taking 7x=t, this equation becomes 7t2-50t+7=0
i.e., (7t-1) (t-7) = 0
By factorization

14. QUADRATIC EQUATIONS
 t =
1
7
and 7 are roots of the above equation
 7x =
1
7
(or) 7x = 7
 The solution set of the given equation is {-1,1}
 x = -1 or x = 1
Bases are equal.
On equating the
powers we get

15. QUADRATIC EQUATIONS
2) Solve
x
x−3
+
x−3
x
=
5
2
Solution On taking
x
x−3
=t, the given equation become
t+
1
t
=
5
2
⇒
t2+1
t
=
5
2
 2t2-5t+2=0
By factorization
1
2
and 2 are the roots of the above equation
 2t2-4t-t+2=0

16. QUADRATIC EQUATIONS
x
x−3
=
1
2
(or)
x
x−3
= 2 ⇒
x
x−3
=
1
4
(or)
x
x−3
= 4
 4x = x-3 (or) x = 4x-12
 x = -1 (or) x = 4
It can be verified that x=-1 and x=4 satisfies the
given equation
 The solution set of the given equation is {-1, 4}

17. QUADRATIC EQUATIONS
3) Solve 2 x+
1
x
𝟐
- 7 x+
1
x
+ 5=0
Solution
On taking x+
1
x
=t, the given equation becomes
 2t2-7t+5=0
 (2t-5)(t-1)=0
t=
5
2
and ‘1’ are roots of above equation
By factorization

18. QUADRATIC EQUATIONS
x +
1
x
=
5
2
(or) x +
1
x
=1
⇒
x2+1
x
=
5
2
(or)
x2+1
x
=1
 2x2-5x+2=0 (or) x2-x+1=0
 2x−1 x−2 =0 (or) x =
1 1−4 1 1
2 1
 x =
1
2
,2 (or) x =
1i 3
2
∵ i2=-1

19. QUADRATIC EQUATIONS
 The solution set of given equation is
1
2
,2,
1+i 3
2
,
1−i 3
2

20. QUADRATIC EQUATIONS
4) Solve 2x4+x3-11x2+x+2=0
Solution
Since x=0 is not a solution of the given equation
 2 x
2
+
1
x𝟐 + x+
1
x
-11 = 0
2x2+x-11+
1
x
+
2
x𝟐 = 0
On dividing the given equation by x2, we get

21. QUADRATIC EQUATIONS
On substituting x+
1
x
=y, then we get
2(y2-2)+y-11 = 0
 2y2+y-15=0
 (2y-5)(y+3)=0
 y=
5
2
or y= -3
 x+
1
x
=
5
2
(or) x+
1
x
= -3
x+
1
x
= y gives
x𝟐+
1
x𝟐= y2-2

22. QUADRATIC EQUATIONS
 2x−1 x−2 =0 (or) x =
− 3 9−4 1 1
2 1
 x =
1
2
,2 (or) x =
−3 5
2
 The solution set of given equation is
1
2
,2,
−3+ 5
2
,
−3− 5
2

x2+1
x
=
5
2
(or)
x2+1
x
= -3
 2x2-5x+2=0 (or) x2+3x+1=0

23. QUADRATIC EQUATIONS
5) Prove that there is a unique pair of consecutive odd
positive integers such that the sum of their squares is
290 and find it.
Solution
Since two consecutive odd integers differ by 2, we
have to prove that there is unique positive odd integer
x, such that
x2+(x+2)2=290  (1)
 2x2+4x-286=0
 x2+2x-143=0

24. QUADRATIC EQUATIONS
 x2+13x-11x-143 = 0
 (x+13)(x-11) = 0
 x = -13 (or)
Hence ‘11’ is the only positive odd integer
satisfying equation (1)
 (11, 13) is the unique pair of integers which
satisfies the given condition

Now (x,x+2) = (11, 13)

x=11
Given that the
number is odd
positive integer.

25. QUADRATIC EQUATIONS
6) Find the number which exceeds its positive square root
by 12
Solution
Let ‘x’ be any such number
Then x = x + 12  (1)
 x -12 = x
Squaring on both sides, we get
(x-12)2 = x  x2-24x+144 = x

26. QUADRATIC EQUATIONS
 x2-25x+144 = 0
 (x-16)(x-9) = 0
 x = 16 and x = 9 are roots of above equation
But x=9 does not satisfy the equation (1) while
x=16 satisfies (1)
 Required number is 16.

27. QUADRATIC EQUATIONS
7) The cost of a piece of cable wire is Rs.35/-. If the
length of the piece of wire is 4 meters more each
meter costs Rs.1/- less, the cost would remain
unchanged. Find the length of the wire.
Solution
Let the length of the piece of wire be ‘l ’ meters and
the cost of each meter be Rs. x/-, then l.x=35(1)

28. QUADRATIC EQUATIONS
(l+4)(x-1) = 35
 lx-l+4x-4 = 35
 35-l+4x-4 = 35
 4x=l+4
Given the length of the piece of wire is 4 meters more
each meter costs Rs.1/- less
 x =
l+4
4
From eq(1)
lx = 35

29. QUADRATIC EQUATIONS
Substitute ‘x’ in (1), we get
l
l+4
4
= 35  l 2+4l-140=0
 (l+14)(l-10)=0
 l = -14 (or) l =10
Since the length cannot be negative,
 l =10 meters

30. QUADRATIC EQUATIONS
8) If x1, x2 are the roots of ax2+bx+c=0 then find the value
of (ax1+b)-2+(ax2+b)-2
Solution x1 is the root of ax2+bx+c=0  ax1
2+bx1+c=0
 x1(ax1+b) = -c 
1
ax1+b
= -
x𝟏
c
 ax𝟏+b −𝟐=
x𝟏
𝟐
c𝟐

1
ax𝟏+b
𝟐
= −
x𝟏
c
𝟐

31. QUADRATIC EQUATIONS
ax𝟏+b −𝟐
+ ax𝟐+b −𝟐
=
x𝟏
𝟐
+x𝟐
𝟐
c𝟐
=
x𝟏+x𝟐
𝟐−2x𝟏x𝟐
c𝟐
=
−b
a
𝟐
−2.
c
a
c𝟐
=
b𝟐
−2ac
a𝟐c𝟐
Now ax𝟏+b −𝟐 + ax𝟐+b −𝟐 =
x𝟏
𝟐
c𝟐 +
x𝟐
𝟐
c𝟐
x1 + x2 =
−b
a
x1x2=
c
a
Similarly ax𝟐+b −𝟐
=
x𝟐
𝟐
c𝟐

32. QUADRATIC EQUATIONS
9(i) If ,  are the roots of ax2+bx+c=0 find the quadratic
equation whose roots are p, p
Solution
,  are the roots of ax2+bx+c=0 we have β =
−b
a
, β =
c
a
ppβ = p β =
− pb
a
p pβ = p2 β =
p𝟐
c
a

33. QUADRATIC EQUATIONS
The equation whose roots are P and P is
x2-(P+P)x+(P)(P) = 0
 x2+
pb
a
x+
p𝟐
c
a
=0
 ax2+ pb x+ p𝟐c =0

34. QUADRATIC EQUATIONS
9(ii) If ,  are the roots of ax2+bx+c=0 find the quadratic
equation whose roots are 2+2,-2+-2
Solution
𝟐+𝟐
+
1
𝟐 +
1

𝟐 = + 𝟐−2 +
𝟐
+
𝟐
𝟐

𝟐
=
b
𝟐
a𝟐 -
2c
a
+
b
𝟐
a𝟐 − 2c
a
c𝟐
a𝟐
=
b
𝟐
−2ac
a𝟐 +
b
𝟐
−2ac
c𝟐
=
b𝟐
−2ac a𝟐+c𝟐
a𝟐c𝟐
Firstly we calculate
sum of the roots,i.e,
We know that
𝟐
+𝟐
=
b𝟐
a𝟐 −
2c
a

35. QUADRATIC EQUATIONS
𝟐+𝟐
×
1
𝟐
+
1
𝟐
=
b𝟐
a𝟐 −
2c
a
𝟐
c𝟐
a𝟐
=
b𝟐
−2ac
𝟐
a𝟐c𝟐
=
𝟐+𝟐 𝟐
𝟐𝟐
=
 +  𝟐 − 2
𝟐
 𝟐
Next, we calculate
product of the roots, i.e,

36. QUADRATIC EQUATIONS
x𝟐 −
(b
𝟐
−2ac)(a𝟐
+c𝟐
) x
a𝟐
c𝟐 +
b
𝟐
−2ac
𝟐
a𝟐
c𝟐 = 0
⇒ a𝟐c𝟐x𝟐 − b𝟐
−2ac a𝟐+c𝟐 x + b𝟐
−2ac
𝟐
= 0
 The equation whose roots are 𝟐+𝟐
,
1
𝟐 +
1

𝟐 is

37. QUADRATIC EQUATIONS
10) Find the quadratic equation for which the sum of the
roots is 1 and the sum of the squares of the roots is 13.
Solution
Let  and  be the roots of the required equation, then
+ =1
2+2 =13  + 𝟐
-2 =13
 𝟏 -2 =13
  = -6

38. QUADRATIC EQUATIONS
Required equation =
 x2 - 1 x -6 = 0
 x2 -x -6 = 0
x2 - + x + = 0

39. QUADRATIC EQUATIONS
11) Find the condition that one root of ax2+bx+c=0 may
be n times the other root.
Solution
Let  and n be the roots of given equation, then
+n = -
b
a
()n =
c
a
Substituting (1) in (2) we get
  =
−b
a n+1
(1)
 n2 =
c
a
(2)

40. QUADRATIC EQUATIONS
n
−b
a 1+n
𝟐
=
c
a 
nb
𝟐
a𝟐
1+n
𝟐 =
c
a
 nb𝟐
= 1+n 𝟐
ac

41. QUADRATIC EQUATIONS
12) If the equation x2+ax+b=0 and x2+cx+d=0 have a
common root and the first equation has equal roots
then prove that 2(b+d)=ac
Solution
Let  be a common root of x2+ax+b=0 and x2+cx+d=0
then
2+a+b=0  (1)
But first equation has equal roots
2+c  +d=0  (2)

42. QUADRATIC EQUATIONS
Then +  = -a
from eq (2)
  =
−a
2
and 2 =b 2+c  +d=0  (2)
 b -
a
2
(c) +d = 0
 2 (b+d) = ac

43. QUADRATIC EQUATIONS
Thank you…