Class 11 Chapter 5 Complex Numbers and Quadratic Equations (Quadratic Equations)Lecture 6.pdf

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Class 11 Chapter 5 Complex Numbers and Quadratic Equations Lecture 6
#nature_of_roots #symmetric_functions #transformations_of_equations
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1. Dr. Pranav Sharma
Maths Learning Centre. Jalandhar.

2. identity: If two expressions are equal for all values of x, then the statement of
equality between the two expressions is called an identity.

If equation (22 — 5A + 6)x? + (A? — 32 + 2)x + (A? — 4) = 0 is satisfied by more
thantwo values of x, find the parameter A.
tf an equation of degree two Is satisfied by more than two values of unknown, then
tt must be an identity. Then, we must have

a? -5A+6=0,427 -344+2=0,47-4=0
214=2,2044= L ZmAi=2,-2
Common value of A which satisfies each condition is A = 2.

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3. (x+b)(x+c) (x+e)(x+a)., (x+a)(x+b)

SRE (b-a)(c-a) * (c-b)(a-b) * (a-c)(b-0 ae iene RES
When x = —a, then LHS of Eq. = ee = 1=RHS
When x = —b, then LHS of Eq. = eet 1=RHS

and when x = —¢, then LHS of ea. = OO = 1 = RHS

Thus, highest power of x occurring in relation of Eq. ts 2 and this relation is
satisfied by three distinct values of x(= —a, —b, —c) . Therefore, it cannot be an
equation and hence it is an identity.

Showthat x? — 5|x| + 2 = 0 is an equation.

Put x = Oin x? — 3|x| +2 =0 = 0? -3|0/+2=2+0

Since, the relation x? — 3|x| +2 = 0 is not satisfied by x = 0. Hence, itis an
equation.

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4. Linear Equation: An equation of the form ax + b = 0 where a,b € Rand a + 0, is
a linear equation.

Quadratic Equation: An equation in which the highest power of the unknown
quantity ts 2, is called a quadratic equation.

1. Purely Quadratic Equation: A quadratic equation in which the term containing
the first degree of the unknown quantity is absent, is called a purely quadratic
equation. Le, ax? + c = 0, where a, c E C anda # 0.

2. Adfected Quadratic Equation: A quadratic equation in which it contains the
terms of first as well as second degrees of the unknown quantity, is called an
adfected (or complete) quadratic equation. Le. ax? + bx+c=0,

where a, b,c € Canda#0,b+0.

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5. Standard Quadratic Equation: An equation of the form ax? + bx +c= 0 (t)
where a, b,c EC and a + O, is called a standard quadratic equation.
The numbers a, b, c are called the coefficients of this equation.

A root of the quadratic Eq. (1) is a complex number &, such that
aad’ +ba+c=0.
D = b? — 4acís the disoriminant of the Ga. (1) and its roots are given by the

~btvD
X = —

following formula. =

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6. atu ts
1. (fa, b,CER and a + 0, then
(t) HD < 0, then Ex. (i) has non-real complex roots.

(ti) FD > 0, then Ea. (t) has real and distinct roots, namely
—b+VD —b-vD
ze am Tenga

X1= and then ax? + bx +.¢ = a(x —x4)(x — x2).(ü)

(tit) tf D = 0, then Eq. (t) has real and equal roots, then X1 = Xz = -2 and then
ax? + bx + c = a(x — x1)?...((0)

To represent the quadratic ax? + bx+cin form Eas. (ti) or (tit), is to expand it

into linear factors.

(iv) tf D = 0, then Ea. (£) has real roots.

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7. (v) tf Dy and Dz be the discriminants of two quadratic equations, then

(a) tf Dy + Dz > 0, then
> At-least one of Dy and D > 0.

(0) tf Da + Dz < 0, then

> At-Least one of Di and Dz < 0.

2. fa, b,c € Q and D is a perfect square of a rational number, the roots are rational
and in case it is not a perfect square, the roots are irrational.

3. tfa, b,c € Rand p + ig is one root of Eq. (I) (q + 0), then the other must be the
conjugate (p- iq) and vice-versa (where, p,q € Rand i= V—1).

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8. 4.1f a,b,c EQ and p+ Va is one root of Eq. (1), then the other must be the
conjugate p — /q and vice-versa (where, p is a rational and /q is a surd).

5.tfa=1and b,c E I and the roots of Eq. (1) are rational numbers, these roots
must be integers.

e.tfa+b+c=0 and a, b, c are rational, 1 is a root of the Eq. (1) and roots of the
Ea. (1) are rational.

7.a2 + b?+c2-ab-bce-ca=2{(a - b)? + (b-c)? + (c—a)*}
= -{(a-b)(b-c)+(b-c)(c-a)+(c-a)(a-b)}

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9. Find all values of the parameter a for which the quadratic equation
(a+ Da? +2(a+ Dx+a-2=0

(£) has two distinct real roots. (it) has no real roots. (tit) has two equal
roots.
By the hypothesis, this equation is quadratic and therefore a + —1 and the
discriminant of this equation,

D=4(a+1)?— 4(a + 1)(a — 2) = 4(a + D(a+1-a+2)=12(a+1)
@) For a > (—1), then D > 0, this equation has two distinct roots.
(ti) For a < (—1), then D < 0, this equation has no roots.
(tit) This equation cannot have two equal roots. Since, D = 0 only for a = —-1 and
this contradicts the hypothesis.

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10. 123 x2-3

solve forx, (5+2V6) +(5-2V6) =
1
(5 +2V6)(5 — 246) = 1 so, (5 — 246) = 20
x2-3 12-3

(5+2V6) ~+(5-2V6) =

+ 23
reduces to (5 + 2/6)" ° + E) =10

>
put (5 +2V6) =t thent++=10
>t -10t+1=0ort= 102/00= 2 = (542,6)
2_

> (5 + 2/6)" 3 _ (64206) =(5+ 218)" 0,22 -3= +1
=> x? —3 = Torx? —3 = 15 x? = 4 oy x? = 2
Hence, x = +2, +V2

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11. Show that if p,q, r and s are real numbers and pr = 2(q +5), thew atleast one of
the equations x” + px + q = 0 and x? +1rx +s = 0 has real roots.

Let Dy and Dz be the discriminants of the given equations x? + px + q = 0 and
x? + rx +s = 0, respectively.

Now, Di + Dz =p?-—4q+r?-4s=p?+r?-4(q+s8) =p? +r? -2pr
Igiven, pr = 2(q + s)]

=(p-r)?>0L:-p and q are real] or Di + Dz > 0

Hence, atleast one of the equations x” + px + q = 0 and x? +1rx +s = 0 has real
roots.

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12. tfa,ß are the roots of the equation (x — a)(x—b) = ¢, #0. Find the roots of the

equation (x — a)(x—- B)+c=0.

Since, à, B ave the roots of (x — a)(x — b) = c or(x-a)(x-b)-c=0,

Then (x LADY) = c (xa) (x — B)
>(x-a)(x-P)+c=(x-a)(x-b)

Hence, roots of (x — a)(x — B) + c = 0 are a, b.

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13. Find all roots of the equation x* + 2x3 — 16x? = 22x +7 = 0, (fone root is
2+ V3.
AU coefficients are real, irrational roots will occur in conjugate pairs.
Hence, another root is 2 — V3.
Product of these roots = (x — 2 — V3)(x — 2 + V3)
=(x-2)-3=x7-4x+1.
On dividing x* + 2x3 — 16x? — 22x +7 by x? — 4x + 1, then the other quadratic
factor is x? + 6x + 7.
Then, the given equation reduce in the form (x? —4x + 1)(x2+ 6x +7) = 0
So,x7+6x+7=0

The = ER - 3 4 V2

Hence, the other roots are 2 — V3, —3 + V2.

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14. Relatiow between Roots and Coefficients
1. Relation between roots and coefficients of quadratic equation: tf roots of the
equation ax? + bx + c = O(a + 0) be real and distinct and a < B, then

_ =bt+vD DO
== B= 2a *
(£) Sum of roots =S =a+B=-=
(6) Product of roots= P = ap ===

a Coefficient of x2"

vary = pie me ND
(tii) Difference of roots= D' = a — B = a

_ _ Coefficient of x

~ Coefficient of x2"
Constant term

alo

2. Formation of an equation with given roots: A quadratic equation whose roots are
a and B, is given by (x — a) (x — B) = O or x? — (a+ B)x+ ap = 0e.
x? — (sum of roots) x + Product of roots = 0

x? —Sx+P=0.

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15. 3. Symmetric function of roots: A function of a and Bis said to be symmetric

function, tf it remains unchanged, when a and B are interchanged.

a? +3a?ß + 308? + B? is a symmetric function of a and B, whereas a? — B3 +

5ap is not a symmetric function of a and B.

In order to find the value of a symmetric function in terms of a+ B, AB and a — B

and also in terms of a, b and c.

b?-2ac
az

© a + B? = (at B)? - 208 = an =

a

A (2) = -22

al”

E b

is) @ + 6? = (a + 8) - 3ap(a+ 8) = (-2) -3(2)(-2)=-(

Pza)
a a E .

la

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16. (wv) a — B? = (a — BY +3aß(a - B) = a +3) (1) - POr

a

2
Wat + pt = (a? +2)? - 202? = (Fe) _

at

(b?-2ac)

(4) at— B* = (a? + B?) (a? — 82) = EA

(vii) a? + p> = (a? + B2)(a? + B?) - a?p?(a + PB)
_ (E zac) (- (13 sabe) i (- 2) = = (25 Sabte+sa?be?)

a a a a as

(Viti) a? — B5 = (a? + B?)(a? — B3) + ap? (a - B)
_ a) Gea) 4. ay (D — D(Dt-3acb?43020?)

al a3 a as

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2 (y _ bt+2a2c2-4ach2

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17. tf one root of the equation x? = ix — (1+ i) = 0, (i=V=1)is1
root.

All coefficients of the given equation are wot real, then other root + 1 — i.
Let other root be A, then sum of roots = ite. 1+ita=i>a = (-1)
Hence, the other root is (—1).

i

find the other

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5
tf one root of the equation x? — V5x — 19 = Ois ons , then find the other root.
All coefficients of the given equation are not rational, then other root + 298
9+V5 -9+,5
Let other root be a, sum of roots = V5 = +a=V5>a= 3
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18. tf the difference between the corresponding roots of the equations x? +ax+b = 0
and x? + bx + a = O(a + b) is the same, find the value of a +b.
Let a, B be the roots of x? + ax + b = 0 and y, 6 be the roots of x? + bx + a = 0,
then given

Va? -4b vb? —4a VD

a Bey = TEN [-«-#-

>a?-4b=b?-4a= (a? — b*) + 4(a— b) = 0 > (a— b)(a+b+4)=0
As, a-b+0 So a+b+4=00a+b=-4.

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19. fa+b+c=0 anda,b,c are rational. Prove that the roots of the equation
(b+c—a)x? + (c+a—b)x+(at+b—c) = 0 are rational.
Given equation ts
(b+c—a)x*+(c+a—b)x+(at+tb—c)=0 (i)
(b+c-a)+(c+ta-b)+(a+tb-c)=a+b+c=0
x = Lis a root of Eq. (£), Let other root of Ea. (t) is a, then
Product of roots = =—

c—c
>1xa=—— [..a+b+c=0]

a= = [rational] Hence, both roots of Ea. (i) are rational.

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20. tf the roots of equation a(b — da? + b(c— a)x+c(a — b) = 0 be equal, prove that
a,b, c are in HP.

Given equation is a(b—c)x? +b(c—a)x+c(a—b)=0 (i)

Here, coefficient of x? + coefficient of x + constant term = 0

Le, a(b—c)+b(c—a)+c(a—b)=0

Then, 1 ts a root of Ea. (i).

Since, its roots are equal.

Therefore, its other root will be also equal to 1.

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= __ c(a-b) Aa Le me
Then, product of roots = 1x 1 = Ea? ab ac=ca-bc
So, b = =
Hence, a, b and.c are in HP.
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21. fais a root of 4x? + 2x — 1 = 0. Prove that 4a? — 3a is the other root.
2 1 1 .

Let other root ts B, then a+ B= rs —,0rB =-,-4 (t)

and so 4a? + 2a — 1 = 0, because ais a root of 4x? + 2x-1=0.

Now, B = 4a? — 3a = a(4a? — 3) = a(1-2a-3) |... 4a? +2a-1=0]
1 1
= 2200 - 2a = —5 (4a") - 2a = —5 (1 - 2a) - 2a|. 4a? +20 —1 = 0]

1 .
q B [from €q. (()1
Hence, 4a? — 3a is the other root.

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22. a, B ave the roots of the equation A(x? — x) + a+ r + Ba tf Aq and Az ave ini
values of A for which the roots a, B are related bys +2 à find the value of +2
The given equation can be written as Ax? — (A- x qi 5= = 0, a, B are the rote 7%

4-1 5

this equation. So, a+ PB == and ap =>

so a, B_4_l,a+p? _4

Bin One Centre anis
2-2 a OFS ao, à

de A ASA U PR rh
aß 5 E 5 52 5

>42 —16A+1= 0 itis a quadratic in A, Let roots be Ay and Az, then
A]

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May Az A _ Artdz)?-2ardg OD
u Az “+ ay M A2 1 = 254
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23. Transformation of Quadratic Equations
Let a, B be the roots of the equation ax? + bx + c = 0, then the equation
(i) whose roots area + k, B + k,is
a(x = k)? + b(x—k) + c = 0 [replace x by (x — k)l

(it) whose roots are a— k, B — k, is
a(x +1)? + b(x +k) + c = 0 replace x by (x + k)l

(it) whose roots are ak, Bk, is ax? + kbx + k?c = 0 [replace x by Gy

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24. (tv) whose roots ave =, £ is ak?x? + bkx +c = 0 [replace x by xk1

(v) whose roots are —a, —B,ís ax? — bx +c =0 [replace x by (-x)]

is ex? + bx+a=0 [replace x by O

Ria

à 1
(vi) whose roots are =, B’

1

A 2 ur 1
zs cx“—bx+a=0 [replace x by (-2)

ae 1
(vit) whose roots are — ao

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25. o leo Bale 2 2h L k
(iii) whose roots are =, gs exe + kbx + k°a = 0 [replace x by Gy

(x) whose roots are pa + q, pB + q, is
a(=) +b (=) +c=0 [replace x by E)

(x) whose roots area”, B",nEN,ís
a(x!/n)? + b(x¥/") +c = 0 [replace x by (x"/®)]

(xt) whose roots are a", BY" ne Nis
a(x")? + b(x") + c = 0 [replace x by (1

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26. tf a, B be the roots of the equation x? — px + q = 0, then find the equation whose

roots ave and I.
p- p-B

= tll
LG TAR AZP E y
So, we replacing x by p — = in the given equation, we get

=D oro) + 9H pe + le 0

; ;
ee ee re

is the required equation whose roots ae and urs oy

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27. fa and Bare the roots of ax? + bx + c = 0, then find the roots of the equation

<
ax? — bx(x — 1) + c(x — 1)? =0. 9
ax? — bx(x- 1) +c(x- 1)? = 0 (i) 5
x V2 x x Y x ®
>a(5) -6(4)te=0 oral) +o(=)+c=0 a
Now, a, B ave the roots of ax? +bx+c=0. El
Lea re, J% n a lathB
Then, a = 1. And B = mile and x= E
® E X ©
Hence, q + pri are the roots of the Eq. (i). E
=
mM
e
=
n
©
=
5
o
a
o
2
i=
=
o
n
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28. “apne
tf a, B be the roots of the equation 3x? + 2x + 1 = 0, then find value of =

a?
5)
tx a= 2
Ea LE
So, replacing x by 7 aa in the given equation, we get
3 1 er Jamon -2x+3=0..()

1+
ttis olear that + ae = and m are the roots of Eq. (

so, (5a) + (ES =2 (4) and ag Gr 7) = 3 Gi)

_a\3 =p\3
at aap nec?

EDS -28-3.3.2=s-18=-10

1+a/\1+B/\1t+a@ 1+8

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29. Roots Under Special Cases

Consider the quadratic equation ax? +bx+c=0 (i)
where a, b,c ER and a # 0. Then, the following hold good:

(© tf roots of Eq. (1) are equal in magnitude but opposite in sign, then sum of roots
ts zero as well as D > 0,í.e.b=00nd D > 0.

(i) tf roots of Eq. (i) are reciprocal to each other, then product of roots is 1 as well as
D=0ie,a=candD=0.

(tit) tf roots of Eq. (t) are of opposite signs, then product of roots < 0 as well as D >
Ote,a>0,c<OandD>Oora<0,c>O0andD>0.

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30. (tv) tf both roots of Eq. (1) are positive, then sum and product 1 roots > 0 as well as
D>0í,,a>0,b<0,Cc>0mdD>00a<0,b>0,c<00madD>0.

(v) tf both roots of Eq. (L) are negative, then sum of roots < 0, product of roots > 0
aswelas D>0ie,a>0,b>0,c>0and D >00ora<0,b<0,c<0and
D20.

(vi) tf greater root in magnitude of Eq. (i) ts positive, then sign of b =sign of
c #sign of a.

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31. (vit) tf greater root in magnitude of Eq. (t) is negative, then sign of a =sign of
b #sign of c.

(viti) tf both roots of Eq. (I) are zero, then b=c = 0.

ge
a

(ix) tf roots of Eq. (0) are 0 and (—2), then c = 0.

0%) If roots of Ea. (i) are Land ©, then à + b+c = 0.

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32. For what values ofm, the equation x24+2(m=1)x +m+5= 0 has (m € R)
(t) roots are equal in magnitude but opposite in sign?

(i) roots ave reciprocals to each other? (tit) roots are opposite in sign?
(tv) both roots are positive? (v) both roots are negative?
(vi) atleast one root is positive? (vit) atleast one root is negative?

Here, a=1,b=2(m-1)andc=m+s
D = b? — 4ac = 4(m - 1)? — 4(m + 5) = 4(m? — 3m — 4)
D = 4(m — 4)(m + 1) and herea=1>0

() b = 0 and D > 0 = 2(m —1) = 0 and 4(m — 4)(m + 1) > 0
>m=1andme (—o, —1) U (4,00) m € @ [null set]

(i)a=candD>0 n=1=m+5and4m-4(m+1)>20
>m=-4andm € (—~,—1] U [4,0) m=-4

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(it) a>0,c<0maAD>0>1>0,m+5<0and 4(m — 4)(m +1) > 0
>m < =5 and m € (-0, -1) U (4,0) m € (—co,-5)

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33. (iv) a>0,b<0,c>0andD=0 0r>1>0,2(m-1)<0,m+5>0
and 4(m — 4)(m +1) >0>m<1,m>-—5andme (—o,—1] U [4,0)
=>me (-5,-1]

(Y a>0,b>0,c>00mdD>0>1>0,2(m-1)>0,m+5>0
and 4(m— 4)(m +1) > 0 = m > 1,m > —5 and m € (-o,—1] U [4, 00)
m € [4, 00)

(vi) Either one root ts positive or both roots are positive
Le., (tii) U (iv) > m € (—~, —5) U (-5,—1]

(Vit) Either one root is negative or both roots are negative
Le, (iii) U (v) > m € (—~, —5) U [4, co)

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34. Condition for Common Roots

1. Only One Root is Common: Consider two quadratic equations ax? +bx+c=0
and a'x2 + b’x+c'=0 [where a, a’ + 0 and ab! - a'b + 0]

Let ar be a common root, then aa? +ba+c=Oandaa?+ba+c=0.

On solving these two equations by cross-multiplication, we have

a? a 1
bc'—b'e caca ab'-a'b
2 be’ -b’c A
From first two relations, we get a =, (1
Lal
and from Last two relations, we get a = eae (i)

lb
From Eas. (i) and (it), we get ——— Hei = ==

ca’-ca arb-abı

> (ab' - a'b) (bc' — b'c) = (ca' - c'a)?

ly

quadratic equations to be common.

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2
b x M el = ke al This is the required condition for one root of two

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35. 2. Both Roots are Common
Let a, B be the common roots of the equations
ax? + bx + c = 0 and a'x? + b'x + c' =0, then

b sentra Jalandha
=-- =- —-5-=-
a+ß = ae (itt)

c c a Ch
=-=-32-=-
and aß A mr; (iv)

Ay i b
From Gas. (iit) and (iv), we get = =
This is the required condition for both roots of two quadratic equations to be
identical.

ale

To find the common root between the two equations, make the same coefficient of x?
in both equations and then subtract of the two equations.

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36. Find the value of A, so that the equations x? ~x—12 = 0 and Ax? +10x+3 = 0

<
may have. one root in common. Also, find the common root. 2
x? -x-12=0=>(x-4)(x+3)=0 sox=4,-3 5
fx = 4 is a common root, then A(4)? + 10(4) +3 = 0 So, À = — E 2
and if x = —3 is a common root, then A(-3)? + 10(—3) + 3 = 0 So, À = 3 S
=
Hence, for À = E, common root is x= 4 and for À = 3, ®
common root is x = —3. ES
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37. tf equations ax? +bx+\¢ = 0, (where a,b, cER and a #0) and x? +2x+ 5 = 0

<
have a common root, then showthat a: b:c = 1: 2:3. 2
Given equations are ax +bx+c=0 (i) and x? + 2x +3 =0 (ii) 5
Clearly, roots of Eq. (ti) are imaginary, since Eas. (£) and (it) have a common root. ®
Therefore, common root must be imaginary and hence both roots will be common. 8
Therefore, Eqs. (1) and (it) are identical, 3
de eR oO we Lids ®
17330 a:b:c=1:2:3 ES
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Maths Learning Centre, Jalandhar https://sites.google.com/view/mathslearningcentre

38. tf a,b, care in GP, show that the equations ax? +2bx+c=0 and

b
dx? + 2ex + f = 0 have a common root, fy, = 7 arein HP.

Given equations areax? +2bx+c=0 (it) And. dx? +2ex+f=0 (i)

Since, a, b, c are in GP.
b? = ac or b = Vac
ve

2
From Ea. (i), ax? + 2Vacx + c = 0 or (Vax + ve) =00rx = Ta
Given Gas. (1) and (it) have a common root.

ES
Hence, x = mite also satisfied Ea. (it), then
c ve d aff
d Re —=r 0>--—+==0

y ()- Va f= a FM c

ur La 0 [-- b = vac]
a bc

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