ML Aggarwal Solution Class 10 Chapter 5 Quadratic Equations in One Variable Exercise 5.4

 Exercise 5.4

Question 1

Find the discriminant of the following equations and hence find the nature of roots:

(i) 3x² – 5x – 2 = 0

(ii) 2x² – 3x + 5 = 0

(iii) 7x² + 8x + 2 = 0

(iv) 3x² + 2x – 1 = 0

(v) 16x² – 40x + 25 = 0

(vi) 2x² + 15x + 30 = 0.

Sol :

(i) 3x² – 5x – 2 = 0

Here a = 3, b = -5, c = -2

D=b24ac=(5)24×3×(2)=25+24=49

∴ Discriminant =49

∵D>0

∴ Roots are real and distinct


(ii) 2x23x+5=0

Here a=2, b=-3, c=5

D=b24ac=(3)24×2×5=9-40=-31

∴ Discriminant =-31

∵D<0

∴ Roots are not real.


(iii) 7x2+8x+2=0

Here, a=7, b=8, c=2

D=b24ac=(8)24×7×2=64-56=8

∵Discriminant =8

∴ D>0

∴ Roots are real and distinct


(iv) 3x2+2x1=0

Here a=3, b=2, c=-1

D=b24ac=(2)24×3×(1)=4+12=16

∴  Discriminant =16

∴ D>0

∴  Roots are real and distinct


(v) 16x240x+25=0

a=16, b=-40, c=25

D=b24ac=(40)24×16×25=1600-1600=0

∴  Discriminant =0

∵ D=0

∴  Roots are real and equal.


(vi) 2x2+15x+30=0

Here, a=2, b=15, c=30

D=b24ac=(15)24×2×30=225-240=-15

∴ Discriminant =-15

∵ D<0

∴ Root are not real.


Question 2

Discuss the nature of the roots of the following quadratic equations :

(i) x² – 4x – 1 = 0

(ii) 3x22x+13=0

(iii) 3x243x+4=0

(iv) x212x+4=0

(v) 2x2+x+1=0

(vi) 23x25x+3=0

Sol :

(i) x² – 4x – 1 = 0

Here a = 1, b = -4, c = -1

D=b24ac=(4)24×1×(1)=16+4=20
∵D>0
Roots are real and distinct

(ii) 3x22x+13=0
Here a=3, b=-2, c=13

D=b24ac=(2)24×3×13=44=0

∵D=0

∴Roots are real and equal


(iii) 3x243x+4=0

Here a=3,b=43,c=4

D=b24ac=(43)24×3×4=48-48=0

∵D=0

∴Roots are real and equal


(iv) x212x+4=0

Here, a=1,b=12,c=4

D=b24ac=(12)4×1×4=1416=634

∵D=0

∴Roots are real and equal


(v) 2x2+x+1=0

Here, a=-2, b=1, c=1

D=b24ac=(1)24×(2)×1=1+8=9

∵D=0

∴Roots are real and equal


(vi) 23x25x+3=0

Here a=23,b=5,c=3

D=b24ac=(5)24×23×3=2524=1

∵D=0

∴Roots are real and equal


Question 3

Find the nature of the roots of the following quadratic equations:

(i) x212x12=0
(ii) x² – 2√3x – 1 = 0 If real roots exist, find them.
Sol :
(i) x212x12=0

Here a=1,b=12,c=12

D=b24ac

=(12)24×1×(12)

=14+2=94

D=94>0

∴Roots are real and unequal


(ii) x223x1=0

Here a=1,b=23,c=1

D=b24ac

=(23)24×1×(1)=12+4=16

∵ D>0

∴ Roots are real and unequal.


Question 4

Without solving the following quadratic equation, find the value of ‘p’ for which the given equations have real and equal roots:

(i) px² – 4x + 3 = 0

(ii) x² + (p – 2)x + p = 0.

Sol :

(i) px² – 4x + 3 = 0

Here a = p, b = -4, c = 3

D=b24ac=(4)24×p×3=16-12 p

∵ The roots are equal 

∵ D=0

b24ac=0

⇒ 16-12p=0

⇒ 12p=16

p=1612=43 p=43


(ii) x2+(p3)x+p=0

Here a=1, b=(p-3), c=p

∵ Equation has real and equal roots 

b24ac=0

(p3)24(1)(p)=0

(p3)24p=0

p2+96p4p=0

p210p+9=0

p29pp+9=0

⇒p(p-9)-1(p-9)=0

⇒(p-1)(p-9)=0

∴ p=1,9


Question 5

Find the value (s) of k for which each of the following quadratic equation has equal roots :

(i) kx² – 4x – 5 = 0

(ii) (k – 4) x² + 2(k – 4) x + 4 = 0

Sol :

(i) kx² – 4x – 5 = 0

Here a = k, b = -4, c = 5

D=b24ac=(4)24×k×(5)=16+20 k

∵ Roots are equal. 

∴ D=0

b24ac=0

∴ 16+20k=0 

⇒20k=-16

k=1620=45

Hence k=45


(ii) (k4)x2+2(k4)x+4=0

Here a=k-4, b=2(k-4), c=4

D=b24ac

=[2(k4)]24×(k4)×4

=4(k2+168k)16(k4)

=4(k28k+16)16(k4)

=4[k28k+164k+16]

=4(k212k+32)

∵ Roots are equal 

∴D=0

4(k212k+32)=0

k212k+32=0

k28k4k+32=0

⇒k(k-8)-4(k-8)=0

⇒(k-8)(k-4)=0

Either k-8=0, then k=8

or k-4=0 then k=4

But k40

k4

k=8


Question 6

Find the value(s) of m for which each of the following quadratic equation has real and equal roots:

(i) (3m + 1)x² + 2(m + 1)x + m = 0

(ii) x² + 2(m – 1) x + (m + 5) = 0

Sol :

(i) (3m + 1)x² + 2(m + 1)x + m = 0

Here a = 3m + 1, b = 2(m + 1), c = m

D=b24ac

=[2(m+1)]24×(3m+1)(m)

=4(m2+2m+1)12m24m

=4m2+8m+412m24m

=8m2+4m+4

∴ Roots are equal. 

∴ D=0

8m2+4m+4=0

2m2m1=0 (Dividing by 4)

2m22m+m1=0

⇒ 2m(m-1)+1(m-1)=0

⇒(m-1)(2 m+1)=0

Either m-1=0, then m=1 or 2m+1=0, then 2 m=-1

m=12


Question 7

Find the values of k for which each of the following quadratic equation has equal roots:

(i) 9x² + kx + 1 = 0

(ii) x² – 2kx + 7k – 12 = 0

Also, find the roots for those values of k in each case.

Sol :

(i) 9x² + kx + 1 = 0

Here a = 9, b = k, c = 1

D=b24ac

=k24×9×1=k236

∵Roots are equal

∴D=0

k236=0

⇒(k+6)(k-6)=0

Either k+6=0, then k=-6

∴k-6=0, then k=6

(a) If k=6 then

9x2+6x+1=0

(3x)2+2×3x×1+(1)2=0

(3x+1)2=0

∵3x+1=0

⇒3x=-1

x=13,13


(b) If k=6, then

9x26x+1=0

(3x)22×3x×1+(1)2=0

(3x1)2=03x1=0

3x=1 

x=13

x=13,13


(ii) x22kx+7k12=0

Here a=1, b=-2 k, c=7, k-12

D=b24ac

=(2k)24×1×(7k12)

=4k24(7k12)

=4k228k+48

∵Roots are equal 

∴D=0

4k228k+48=0

k27k+12=0

k23k4k+12=0

k(k-3)-4(k-3)=0

(k-3)(k-4)=0

Either k-3=0, then  k=3

or k-4=0, then k=4


(a) If k=3, then

x=b±D2a=4k±02×1=4×32=122=6

x=6,6

(b) If k=4, then

x=b±D2a=(2×4)±02×1=+82=4

∴x=4,4


Question 8

Find the value(s) of p for which the quadratic equation (2p + 1)x² – (7p + 2)x + (7p – 3) = 0 has equal roots. Also find these roots.

Sol :

The quadratic equation given is (2p+1)x²–(7p+2)x+(7p–3)=0

Comparing with ax² + bx + c = 0, we have

a=2 p+1, b=-(7 p+2), c=(7 p-3)
D=b24ac
0=[(7p+2)]24(2p+1)
(7 p-3)
0=49p2+4+28p4(14p26p+7p3)
0=49p2+4+28p56p24p+12
0=7p2+24p+16
0=7p2+28p4p+16
0=-7 p(p-4)-4(p-4)
0=(-7 p-4)(p-4)
-7 p-4=0 or p-4=0
Hence, the value of p=47 or p=4


Question 9

If – 5 is a root of the quadratic equation 2x² + px – 15 = 0 and the quadratic equation p(x² + x) + k = 0 has equal roots, find the value of k.

Sol :

-5 is a root of the quadratic equation

2x² + px – 15 = 0, then

⇒ 2(5)² – p( -5) – 15 = 0

⇒ 50 – 5p – 15 = 0

⇒ 35 – 5p = 0

⇒ 5 p=35
p=355=7

p(x2+x)+k=0 has equal roots px2+px+k=0

7x2+7x+k=0

Here, a=7, b=7, c=k

b24ac=(7)24×7×k

=49-28 k

∵ Roots are equal 

b24ac=0

⇒49-28 k=0

⇒ 28 k=49 

k=4928=74

k=74


Question 10

Find the value(s) of p for which the equation 2x² + 3x + p = 0 has real roots.

Sol :

2x² + 3x + p = 0

Here, a = 2, b = 3, c = p

b24ac=(3)24×2×p
=9-8 p

∵ Roots are real 

b24ac0

9-8p≥0

9≥8p

8p≤9

p98


Question 11

Find the least positive value of k for which the equation x² + kx + 4 = 0 has real roots.

Sol :

x² + kx + 4 = 0

Here, a = 1, b = k, c = 4

b24ac=k24×1×4
=k216
∵Roots are real and positive
k2160k216
k≥4
k=4

Question 12

Find the values of p for which the equation 3x² – px + 5 = 0 has real roots.

Sol :

3x² – px + 5 = 0

Here, a = 3, b = -p, c = 5

b24ac=(p)24×3×5
=p260
∴Roots are real
b24ac0
p2600p260
p±60=±215
p215 or p215

Comments

Popular posts from this blog

ML Aggarwal Solution Class 10 Chapter 15 Circles Exercise 15.1

ML Aggarwal Solution Class 9 Chapter 20 Statistics Exercise 20.2

ML Aggarwal Solution Class 9 Chapter 3 Expansions Exercise 3.2