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8/13/2019 Qudratic eqn practice for cbse11
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Quadratic EquationsQuadratic Equations
MATHEMATICS
Notes
MODULE - I
Al gebra
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2
QUADRATIC EQUATIONS
Recall that an algebraic equation of the second degree is written in general form as2 0, 0ax bx c a+ + =
It is called a quadratic equation inx. The coefficient a is the first or leading coefficient, bis the second or middle coefficient and c is the constant term (or third coefficient).
For example, 7x+ 2x+ 5 = 0,2
5x+
2
1x+ 1 = 0,
3xx= 0, x+2
1= 0, 2x+ 7x= 0, are all quadratic equations.
In this lesson we will discuss how to solve quadratic equations with real and complex
coefficients and establish relation between roots and coefficients. We will also find cube
roots of unity and use these in solving problems.
OBJECTIVES
After studying this lesson, you will be able to:
solve a quadratic equation with real coefficients by factorization and by using quadraticformula;
find relationship between roots and coefficients;
form a quadratic equation when roots are given; and
find cube roots of unity.
EXPECTED BACKGROUND KNOWLEDGE
Real numbers
Quadratic Equations with real coefficients.
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2.1 ROOTS OF A QUADRATIC EQUATION
The value which when substituted for the variable in an equation, satisfies it, is called a root
(or solution) of the equation.
Ifbe one of the roots of the quadratic equationthen, 2 0, 0ax bx c a+ + = ... (i)
2 0a b c + + =In other words,x is a factor of the quadratic equation (i)
In particular, consider a quadratic equation x2+x 6 = 0 ...(ii)
If we substitute x = 2 in (ii), we get
L.H.S = 22+ 2 6 = 0
L.H.S = R.H.S.
Again put x =3 in (ii), we get
L.H.S. = ( 3)23 6 = 0
L.H.S = R.H.S.
Again put x =1 in (ii) ,we get
L.H.S = ( 1)2+ (1) 6 = 6 0 = R.H.S.
x= 2 andx=3 are the only values ofxwhich satisfy the quadratic equation (ii)
There are no other values which satisfy (ii)
x= 2,x=3 are the only two roots of the quadratic equation (ii)
Note: If, be two roots of the quadratic equation
ax 2+ bx + c = 0, a 0 ...(A)then (x ) and (x ) will be the factors of (A). The given quadraticequation can be written in terms of these factors as (x ) (x ) = 0
2.2 SOLVING QUADRATIC EQUATION BY FACTORIZATION
Recall that you have learnt how to factorize quadratic polynomial of the form
( ) 2 , 0,p x ax bx c a= + + by splitting the middle term and taking the common factors.
Same method can be applied while solving quadratic equation by factorization.
If xpq and x rs are two factors of the quadratic equation
ax2+ bx + c = 0 ,a 0 then (xpq )(x rs ) = 0
either x =pq or, x =
r
s
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The roots of the quadratic equation ax2+ bx + c = 0 arepq ,
r
s
Example 2.1 Using factorization method, solve the following quadratic equation :
6x 2+ 5x6 = 0
Solution: The given quadratic equation is
6x 2+ 5x6 = 0 ... (i)
Splitting the middle term, we have
6x 2+ 9x 4x 6 = 0
or, 3x (2x + 3) 2 (2x + 3) = 0
or, (2x + 3)(3x 2) = 0
Either 2x + 3 = 0 x = 23
or, 3x 2 = 0 x =3
2
Two roots of the given quadratic equation are2
3,
3
2
Example 2.2 Using factorization method,solve the following quadratic equation:
23 x2
+ 7x 23 = 0Solution: Splitting the middle term, we have
23 x2 + 9x 2x 23 = 0
or, 3x ( 2 x + 3) 2 ( 2 x + 3) = 0
or, ( 2 x + 3)(3x 2 ) = 0
Either 2 x + 3 = 0 x =2
3
or, 3x 2 = 0 x =3
2
Two roots of the given quadratic equation are2
3,
3
2
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Quadratic EquationsQuadratic Equations
Example 2.3 Using factorization method, solve the following quadratic equation:
(a + b)2x2+ 6 (a2 b2) x + 9 (a b)2= 0
Solution: The given quadratic equation is
(a + b)2
x2
+ 6 (a2
b2
) x + 9 (a b)2
= 0Splitting the middle term, we have
(a + b)2x2+ 3(a2 b2) x + 3(a2 b2) x + 9 (a b)2 = 0
or, (a + b)x {(a + b) x + 3 (ab) } + 3 (a b) {(a +b) x + 3 (a b) } = 0
or, {(a + b) x + 3 (ab) } {(a + b) x + 3 (a b) } = 0
either (a + b) x + 3 (a b) =0 x =ba
)ba(3
+
=ba
)ab(3
+
or, (a + b) x + 3 (a b) =0 x =ba
)ba(3
+ =
ba
)ab(3
+
The equal roots of the given quadratic equation areba
)ab(3
+
,ba
)ab(3
+
Alternative Method
The given quadratic equation is
(a + b)2x2+ 6(a2 b2) x + 9(a b)2= 0
This can be rewritten as
{(a + b) x}2+ 2 .(a + b)x . 3 (a b) + {3(a b)}2= 0
or, { (a + b)x + 3(a b) }2= 0
or, x =ba
)ba(3
+
=ba
)ab(3
+
The quadratic equation have equal rootsba
)ab(3
+
,ba
)ab(3
+
CHECK YOUR PROGRESS 2.1
1. Solve each of the following quadratic equations by factorization method:
(i) 3 x2 + 10x + 8 3 = 0 (ii) x
2 2ax + a2 b = 0
(iii) x2+
ab
c
c
abx 1 = 0 (iv) x2 4 2 x + 6 = 0
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2.3 SOLVING QUADRATIC EQUATION BY QUADRATIC
FORMULA
Recall the solution of a standard quadratic equation
a
x2
+ bx+ c = 0, a 0 by the Method of Completing SquaresRoots of the above quadratic equation are given by
x1=
a2
ac4bb 2 +
and x2=
a2
ac4bb 2
=a2
Db+, =
a2
Db
where D = b2 4ac is called the discriminant of the quadratic equation.
For a quadratic equation 2 0, 0ax bx c a+ + = if
(i) D>0, the equation will have two real and unequal roots
(ii) D=0, the equation will have two real and equal roots and both roots are
equal to b
2a
(iii) D0.
The equation will have two real and unequal roots
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Verification:By quadratic formula, we have
x =2
419
The two roots are2
419+,
2
419which are real and unequal.
(ii) The given quadratic equation is 9 y2 6 2 y + 2 = 0
Here, D = b2 4 ac
= ( 6 2 )2 4.9.2
= 72 72 = 0
The equation will have two real and equal roots.
Verification:By quadratic formula, we have
y =9.2
026 =
3
2
The two equal roots are3
2,
3
2.
(iii) The given quadratic equation is 2 t2 3t + 3 2 = 0
Here, D = ( 3)2 4. 2 .3 2
=15< 0
The equation will have two conjugate complex roots.
Verification:By quadratic formula, we have
t =22
153
3 15,
2 2
i= where 1i =
Two conjugate complex roots are22
i153+,
22
i153
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Example 2.5 Prove that the quadratic equation x2+ py 1 = 0 has real and distinct roots
for all real values of p.
Solution: Here, D = p2+ 4 which is always positive for all real values of p.
The quadratic equation will have real and distinct roots for all real values of p.
Example 2.6 For what value of k the quadratice equation
(4k+ 1) x2+ (k + 1) x + 1 = 0 will have equal roots ?
Solution: The given quadratic equation is
(4k + 1)x2+ (k + 1) x + 1 = 0
Here, D = (k + 1)2 4.(4k + 1).1
For equal roots, D = 0
(k + 1)2 4 (4k + 1) = 0
k2
14k 3 = 0
k =2
1219614 +
or k =2
20814
= 7 2 13 or 7 + 2 13 , 7 2 13
which are the required values of k.
Example 2.7 Prove that the roots of the equation
x2(a2+ b2) + 2x (ac+ bd) + (c2+ d2) = 0 are imaginary. But if ad = bc,
roots are real and equal.
Solution: The given equation is x2 (a2+ b2) + 2x (ac + bd) + (c2+ d2) = 0
Discriminant = 4 (ac + bd)2 4 (a2+ b2) (c2+ d2)
= 8 abcd 4(a2d2+ b2c2)
= 4 ( 2 abcd + a2d2+ b2c2)
= 4 (ad bc) 2
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CHECK YOUR PROGRESS 2.2
1. Solve each of the following quadratic equation by quadratic formula:
(i) 2x2 3 x + 3 = 0 (ii) 2 2 1 0x x + =
(iii) 24 5 3 0x x + = (iv) 23 2 5 0x x+ + =
2. For what values of k will the equation
y2 2 (1 + 2k) y + 3 + 2k = 0 have equal roots ?
3. Show that the roots of the equation
(x a) (x b) + (x b) (x c) + (x c) (x a) = 0 are always real and they can not be
equal unless a = b = c.
2.4 RELATION BETWEEN ROOTS AND COEFFICIENTS OF
A QUADRATIC EQUATION
You have learnt that, the roots of a quadratic equation ax2+ bx + c = 0, a 0
area2
ac4bb 2 +
anda2
ac4bb 2
Let = a2
ac4bb 2 +
...(i) and = a2
ac4bb 2
... (ii)
Adding (i) and (ii), we have
+ =a2
b2=
a
b
Sum of the roots= 2coefficient of x b
coefficient of x a= ... (iii)
= 2
22
a4
ac)4(bb +
= 2a4
ac4
=a
c
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Product of the roots= 2constant term
coefficient of
c
x a= ...(iv)
(iii) and (iv) are the required relationships between roots and coefficients of a given quadratic
equation. These relationships helps to find out a quadratic equation when two roots aregiven.
Example 2.8 If,, are the roots of the equation 3x2 5x + 9 = 0 find the value of:
(a) 2+ 2 (b) 21
+ 2
1
Solution: (a) It is given that , are the roots of the quadratic equation 3x2 5x +9 = 0.
5
3 + = ... (i)
and9
33
= = ... (ii)
Now, ( )22 2
2 + = +
=
2
3
5
2.3 [By (i) and (ii)]
=9
29
The required value is9
29
(b) Now, 21
+ 2
1
= 22
22
+
=9
929
[By (i) and (ii)]
=81
29
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Example 2.9 If , are the roots of the equation 3y2+ 4y + 1 = 0, form a quadratic
equation whose roots are 2, 2
Solution: It is given that, are two roots of the quadratic equation 3y2+ 4y + 1 = 0.
Sum of the roots
i.e., + = 2yoftcoefficienyoftcoefficien
=3
4... (i)
Product of the roots i.e., = 2yoftcoefficien
termttancons
=3
1... (ii)
Now, 2+ 2 = (+ )2 2
=
2
3
4
2.
3
1[ By (i) and (ii)]
=9
16
3
2
=9
10
and 2 2 = ( )2=9
1[By (i) ]
The required quadratic equation is y2 (2+ 2)y+2 2 = 0
or, y2 9
10y +
9
1= 0
or, 9y2 10y + 1 = 0
Example 2.10 If one root of the equation 2 0, 0ax bx c a+ + = be the square ofthe other, prove that b3+ ac2+ a2c = 3abc
Solution:Let , 2be two roots of the equation ax2+ bx + c = 0.
+2=a
b... (i)
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and .2=a
c
i.e., 3=a
c. ... (ii)
From (i) we have
(+ 1) = a
b
or, { }3
1)( + =3
a
b
= 3
3
a
b
or, 3(3+ 32+ 3+1) = 33
a
b
or,a
c
+
+ 1
a
b3
a
c= 3
3
a
b... [ By (i) and (ii)]
or, 2
2c
a 2
bc3
a+
a
c= 3
3
a
b
or, ac2 3abc + a2c =b3
or, b3
+ ac2
+ a2
c = 3abcwhich is the required result.
Example 2.11 Find the condition that the roots of the equation ax2+ bx + c = 0 are in the
ratio m : n
Solution: Let mand nbe the roots of the equation ax2+bx + c = 0
Now, m+ n= a
b... (i)
and mn2 =a
c... (ii)
From (i) we have, (m + n ) = a
b
or, 2(m + n)2= 2
2
a
b
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or,a
c(m + n)2= mn
2
2
a
b[ By (ii)]
or, ac (m+n)2= mn b2
which is the required condition
CHECK YOUR PROGRESS 2.3
1. If , are the roots of the equation ay2+ by + c = 0 then find the value of :
(i) 21
+ 2
1
(ii) 4
1
+ 4
1
2. If
,
are the roots of the equation 5x2 6x + 3 = 0, form a quadratic equation
whose roots are:
(i) 2, 2 (ii) 3 , 3
3. If the roots of the equation ay2+ by + c = 0 be in the ratio 3:4, prove that
12b2= 49 ac
4. Find the condition that one root of the quadratic equation px2 qx + p = 0
may be 1 more than the other.
2.5 SOLUTION OF A QUADRATIC EQUATION WHEN D < 0
Let us consider the following quadratic equation:
(a) Solve for t :
t2+ 3t + 4 = 0
t =2
1693 =
2
73
Here, D= 7 < 0
The roots are2
73 +and
2
73
or,2
i73 +,
2
i73
Thus, the roots are complex and conjugate.
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(b) Solve for y :
3y2+ 5 y 2 = 0
y = )3(2)2).(3(455
or y =6
195
Here, D = 19 < 0
The roots are5 19 5 19
,6 6
i i +
Here, also roots are complex and conjugate. From the above examples , we can make thefollowing conclusions:
(i) D < 0 in both the cases
(ii) Roots are complex and conjugate to each other.
Is it always true that complex roots occur in conjugate pairs ?
Let us form a quadratic equation whose roots are
2 + 3i and 4 5i
The equation will be {x (2 + 3i)} {x (4 5i)} = 0
or, x2 (2 + 3i)x (4 5i)x + (2 + 3i) (4 5i) = 0
or, x2+ (6 + 2i)x + 23 + 2i = 0
which is an equation with complex coefficients.
Note :If the quadratic equation has two complex roots, which are not conjugate
of each other, the quadratic equation is an equation with complex coefficients.
2.6 CUBE ROOTS OF UNITY
Let us consider an equation of degree 3 or more. Any equation of degree 3 can be expressed
as a product of a linear and quadratic equation.
The simplest situation that comes for our consideration is
x 1 = 0 ...(i)
x1 = (x1) (x +x+ 1) = 0
either x1 = 0 or, x +x+ 1 = 0
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or, x = 1 or, x =2
i31
Roots are 1, 2i3
2
1
+ and 2i3
2
1
These are called cube roots of unity.
Do you see any relationship between two non-real roots of unity obtained above ?
Let us try to find the relationship between them
Let w =2
i3
2
1+
Squaring both sides, we have
w =
2
2
i3
2
1
+
= )i32i31(4
1 2 +
= )i3231(4
1
=4
i322
=( )
4
i312 +=
( )2
i31+
w =2
i3
2
1 = other complex root.
We denote cube roots of unity as 1, w, w
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Roots are1,2
i3
2
1 and
2
i3
2
1+ which can also be written as1 ,w andw2
Therefore, cube roots of1 are1 ,w andw2
In general, roots of any cubic equation of the form x 3= a3 would be a , aw and aw2
Example 2.12 If 1, w and w are cube roots of unity, prove that
(a) 1 + w2+ w7= 0
(b) (1 w + w2) (1 + w w2) = 4
(c) (1 + w)3 (1 + w2)3= 0
(d) (1 w + w2
)3
=8 and (1 + w w2
)3
= 8Solution:
(a) 1 + w2+ w7= 0
L.H.S = 1 + w2+ (w3)2. w
= 1 + w2+ w [ since w3=1]
= 0 [ since 1+ w + w2= 0]
= R.H.S
L.H.S = R.H.S
(b) (1w + w2) (1 + w w2) = 4
L.H.S = (1w + w2) (1 + w w2)
= (1 + w2 w) (1 + w w2)
(since 1 + w2=w and 1 + w =w2)
= ( ww) (w2w2)= ( 2w) ( 2w2)
= 4w3
= 4.1 = 4 =R.H.S
L.H.S = R.H.S
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(c) (1 + w)3 (1 + w2)3= 0
L.H.S = (1 + w)3 (1 + w2)3
= ( w2)3 (w)3 ( 1 + w =w2
= w6
+ w3
and 1 + w2
=w)= (w3)2+ 1
= (1)2+ 1
= 0 =R.H.S
L.H.S = R.H.S
(d) (1 w + w2)3 =8 and (1 + w w2)3=8
Case I :L.H.S = (1w + w2)3
= (1 + w2
w)3
= ( w w)3
= ( 2w)3
= 8w3
= 8 = R.H.S
L.H.S = R.H.S
Case II : L.H.S = (1 + ww)
= ( w2
w2
)3
= ( 2w2)3
= 8w6
= 8(w3)2
= 8 = R.H.S
L.H.S = R.H.S
CHECK YOUR PROGRESS 2.4
1. Solve each of the following cubic equations:
(i) x3= 27 (ii) x3=27
(iii) x3 = 64 (iv) x3=64
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2. If 1, w, w2are cube roots of unity, show that
(i) (1 + w) (1 + w2) (1 + w4) (1 + w8) = 1
(ii) (1 w) (1 w2
) (1 w
4
) (1 w
5
) = 9
(iii) (1 + w)4+ (1 + w2)4=1
(iv) (1 + w3)3= 8
(v) (1 w + w2)6= (1 + w w2)6= 64
(vi) (1 + w)16+ w = (1 + w2)16+ w2=1
LET US SUM UP
Roots of the quadratic equation ax2+ bx + c = 0 are complex and conjugate of eachother, when D < 0.
If, be the roots of the quadratic equation
ax2+ bx + c = 0 then + =b
aand =
a
c
Cube roots of unity are 1, w, w2
where wi
= +1
2
3
2and w
i2 = 1
2
3
2
Sum of cube roots of unity is zero i.e., 1+ w + w2=0
Product of cube roots of unity is 1 i.e., w3= 1
Complex roots w and w2are conjugate to each other.
In general roots of any cubic equation of the form
x3= a3 would be a , aw and aw2
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TERMINAL EXERCISE
SUPPORTIVE WEB SITES
http://www.wikipedia.org
http://mathworld.wolfram.com
1. Show that the roots of the equation
2(a2 + b2)x2 + 2(a + b)x + 1=0 are imaginary,when a b
2. Show that the roots of the equation
bx2 + ( b c)x = c + a b are always real if those of
ax2 + b( 2x + 1) = 0 are imaginary.
3. If , be the roots of the equation 2x2 6x + 5 = 0 , find the equation whose
roots are:
(i)
,
(ii) +
1, +
1(iii) 2+2, 2
1
+ 2
1
4. If 1, w and w2are cube roots of unity , prove that
(a) (2 w) (2 w2) (2 w10) (2 w11) = 49
(b) ( x y) (xw y) ) (xw2 y) = x3 y3
5. If x = a + b , y = aw + bw2and z = aw2+ bw , then prove that
(a) x2+ y2+ z2= 6 ab (b) x y z = a3+ b3
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ANSWERS
CHECK YOUR PROGRESS 2.1
1. (i)
2 34
3, (ii) a b a + b,
(iii)ab
c,
c
ab (iv) 2,23
CHECK YOUR PROGRESS 2.2
1. (i)4
i153 (ii)2
i1
(iii)8
i435(iv)
6
i582
2. 1,2
1
CHECK YOUR PROGRESS 2.3
1. (i) 2
2
c
2acb (ii)
4
2222
c
c2a2ac)b(
2. (i) 25x2 6x + 9 = 0 (ii) 625x2 90x + 81 = 0
4. q2 5p2= 0
CHECK YOUR PROGRESS 2.4
1. (i) 3, 3w, 3w2 (ii) 3,3w ,3w2
(iii) 4, 4w, 4w2 (iv)
4,
4w ,
4w2
TERMINAL EXERCISE
3. (i) 5x2 8x + 5 = 0 (ii) 10x2 42x + 49 = 0 (iii) 25x2 116x + 64 = 0
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