# Class 11 RD Sharma Solutions – Chapter 14 Quadratic Equations – Exercise 14.1 | Set 2

• Last Updated : 21 Dec, 2020

### Question 14. 27x2 – 10x + 1 = 0

Solution:

Comparing the equation with,

ax2+bx+c=0

we get ,a=27,b=-10,c=1

Using Discriminant Method,

D= (b2-4ac)

D= ( (-10)2– 4*27*1)

D= ( 100-108)

√D= √(-8)

√D= 2√2 i

So, roots will be,

R1= (-(-10)+ 2√2 i )/(2*27) and  R2= (-(-10) – 2√2i )/(2*27)

Hence, R1= (5+√2 i)/27 and R2= (5-√2 i)/27.

### Question 15. 17x2 + 28x + 12 = 0

Solution:

Comparing the equation with,

ax2 + bx + c = 0

We get, a=17,b=28,c=12

Using Discriminant Method,

D = (b2-4ac)

D = ((28)2– 4*17*12)

D= (784-816)

√D= √(-32)

√D=4√2 i

So, roots will be,

R1= (-(28)+ 4√2 i)/(2*17) and  R2= (-(28) – 4√2 i)/(2*17)

Hence, R1= (-14+2√2 i)/17 and R2= (-14-2√2 i)/17.

### Question 16. 21x2 – 28x + 10 = 0

Solution:

Comparing the equation with,

ax2+bx+c=0

We get, a=21,b=-28,c=10

Using Discriminant Method,

D= (b2 -4ac)

D= ((-28)2– 4*21*10)

D= (784-840)

√D= √(-56)

√D=2√14 i

So, roots will be,

R1= (-(-28)+ 2√14 i)/(2*21) and  R2= (-(-28)-2√14 i )/(2*21)

Hence, R1= 2/3+ √14 i/ 21 and R2= 2/3 – √14 i/21.

### Question 17. 8x2 – 9x + 3 = 0

Solution:

Comparing the equation with,

ax2+bx+c=0

We get, a=8,b=-9,c=3

Using Discriminant Method,

D= (b2-4ac)

D= ((-9)2 – 4*8*3)

D= (81-96)

√D= √(-15)

√D=√15 i

So, roots will be,

R1= (-(-9)+√15 i)/(2*8) and R2= (-(-9) – √15 i)/(2*8)

Hence, R1= (9+√15 i)/16 and R2= (9-√15 i)/16.

### Question 18. 13x2 + 7x + 1 = 0

Solution:

Comparing the equation with,

ax2+bx+c=0

We get, a = 13, b = 7,c=1

Using Discriminant Method,

D= (b2-4ac)

D= ((7)2 – 4*13*1)

D= (49-52)

√D= √(-3)

√D=√3 i

So, roots will be,

R1= (-(7)+√3 i)/(2*13) and R2= (-(7) – √3 i)/(2*13)

Hence, R1= (-7+√3 i)/26 and R2= (-7-√3 i)/26.

### Question 19. 2x2 + x + 1 = 0

Solution:

Comparing the equation with ,

ax2+bx+c=0

We get, a=2,b=1,c=1

Using Discriminant Method,

D= (b2-4ac)

D= ((1)2– 4*2*1)

D= (1-8)

√D= √(-7)

√D=√7 i

So, roots will be,

R1= (-(1)+√7 i)/(2*2) and  R2= (-(1) – √7i)/(2*2)

Hence, R1= (-1+√7 i)/4 and R2= (-1-√7 i)/4.

### Question 20. √3x2 – √2x + 3√3 = 0

Solution:

Comparing the equation with,

ax2+bx+c=0

We get, a=√3,b=√2,c=3√3

Using Discriminant Method,

D= (b2-4ac)

D= ((√2)2– 4*√3*3√3)

D= (2-36)

√D= √(-34)

√D=√34 i

So, roots will be,

R1= (-(√2)+√34 i)/(2*√3) and R2= (-(√2) – √34i)/(2*√3)

Hence, R1= (-√2+√34 i)/(2√3) and R2= (-√2-√34 i)/(2√3).

### Question 21. √2x2 + x + √2 = 0

Solution:

Comparing the equation with,

ax2+bx+c=0

We get, a=√2,b=1,c=√2

Using Discriminant Method,

D= (b2-4ac)

D= ((1)2– 4*√2*√2)

D= (1-8)

√D= √(-7)

√D=√7 i

So, roots will be,

R1= (-(1)+√7 i)/(2*√2) and R2= (-(1) – √7 i)/(2*√2)

Hence, R1= (-1+√7 i)/(2√2) and R2 = (-1-√7 i)/(2√2).

### Question 22. x2 + x + (1/√2) = 0

Solution:

Comparing the equation with,

ax2+bx+c=0

We get, a=1,b=1,c=1/√2

Using Discriminant Method,

D= (b2-4ac)

D= ((1)2 – 4*1*(1/√2))

D= (1-2√2)

√D= √(-(2√2-1))

√D=√(2√2-1) i

So, roots will be,

R1= (-(1)+√(2√2-1) i)/(2) and R2= (-(1) – √(2√2-1) i)/(2)

Hence, R1= (-1+√(2√2-1) i)/(2) and R2= (-1-√(2√2-1) i)/(2).

### Question 23. x2 + (1/√2)x  + 1 = 0

Solution:

Comparing the equation with ,

ax2+bx+c=0

we get ,a=1,b=1/√2,c=1

Using Discriminant Method,

D= (b2-4ac)

D= ( (1/√2)2– 4*1*1)

D= (1/2-4)

√D= √(-7/2)

√D=√(7/2) i

So, roots will be,

R1= (-(1/√2)+√(7/2)i)/2 and R2= (-(1/√2) – √(7/2)i)/2

Hence, R1= (-1+√7i)/(2√2) and R2= (-1-√7i)/(2√2).

### Question 24. √5x2 + x + √5 = 0

Solution:

Comparing the equation with,

ax2+bx+c=0

We get, a=√5,b=1,c=√5

Using Discriminant Method,

D= (b2-4ac)

D= ( (1)2– 4*√5*√5)

D= (1-20)

√D= √(-19)

√D=√19 i

So, roots will be,

R1= (-(1)+√(19)i)/(2*√5) and R2 = (-(1)-√(19) i)/(2*√5)

Hence, R1= (-1+√19i)/(2√5) and R2 = (-1-√19i)/(2√5).

### Question 25. -x2 + x – 2 = 0

Solution:

Comparing the equation with,

ax2+bx+c=0

We get, a=-1,b=1,c=-2

Using Discriminant Method,

D= (b2-4ac)

D= ((1)2– 4*-1*-2)

D= (1-8)

√D= √(-7)

√D=√7 i

So, roots will be,

R1= (-(1)+√(7)i)/(2*-1) and R2= (-(1)-√(7) i)/(2*-1)

Hence, R1= (-1+√7 i)/(-2) and R2= (-1-√7 i)/(-2).

### Question 26. x2 – 2x + 3/2 = 0

Solution:

Comparing the equation with,

ax2+bx+c=0

We get, a=1,b=-2,c=3/2

Using Discriminant Method,

D= (b2-4ac)

D= ((-2)2 – (4*1*3/2))

D= (4-6)

√D= √(-2)

√D=√2 i

So, roots will be,

R1= (-(-2)+√(2)i)/(2) and R2= (-(-2)-√(2) i)/(2)

Hence, R1= (1+i/√2) and R2= (1-i/√2).

### Question 27. 3x2 – 4x + 20/3 = 0

Solution:

Comparing the equation with,

ax2+bx+c=0

We get, a=3,b=-4,c=20/3

Using Discriminant Method,

D= (b2-4ac)

D= ((-4)2 – (4*3*20/3))

D= (16-80)

√D= √(-64)

√D=8 i

So, roots will be,

R1= (-(-4)+(8)i)/(2*3) and R2= (-(-4)-(8)i)/(2*3)

Hence, R1= (2+4i)/3 and R2= (2-4i)/3.

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