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

### 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)

âˆšD= âˆš(-8)

âˆšD= 2âˆš2 i

So, roots will be,

R1= (-(-10)+ 2âˆš2 i )/(2*27) and  R2= (-(-10) – 2âˆš2i )/(2*27)

Hence, R1= (5+âˆš2 i)/27 and R2= (5-âˆš2 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)

âˆšD= âˆš(-32)

âˆšD=4âˆš2 i

So, roots will be,

R1= (-(28)+ 4âˆš2 i)/(2*17) and  R2= (-(28) – 4âˆš2 i)/(2*17)

Hence, R1= (-14+2âˆš2 i)/17 and R2= (-14-2âˆš2 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)

âˆšD= âˆš(-56)

âˆšD=2âˆš14 i

So, roots will be,

R1= (-(-28)+ 2âˆš14 i)/(2*21) and  R2= (-(-28)-2âˆš14 i )/(2*21)

Hence, R1= 2/3+ âˆš14 i/ 21 and R2= 2/3 – âˆš14 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)

âˆšD= âˆš(-15)

âˆšD=âˆš15 i

So, roots will be,

R1= (-(-9)+âˆš15 i)/(2*8) and R2= (-(-9) – âˆš15 i)/(2*8)

Hence, R1= (9+âˆš15 i)/16 and R2= (9-âˆš15 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)

âˆšD= âˆš(-3)

âˆšD=âˆš3 i

So, roots will be,

R1= (-(7)+âˆš3 i)/(2*13) and R2= (-(7) – âˆš3 i)/(2*13)

Hence, R1= (-7+âˆš3 i)/26 and R2= (-7-âˆš3 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)

âˆšD= âˆš(-7)

âˆšD=âˆš7 i

So, roots will be,

R1= (-(1)+âˆš7 i)/(2*2) and  R2= (-(1) – âˆš7i)/(2*2)

Hence, R1= (-1+âˆš7 i)/4 and R2= (-1-âˆš7 i)/4.

### Question 20. âˆš3x2 – âˆš2x + 3âˆš3 = 0

Solution:

Comparing the equation with,

ax2+bx+c=0

We get, a=âˆš3,b=âˆš2,c=3âˆš3

Using Discriminant Method,

D= (b2-4ac)

D= ((âˆš2)2– 4*âˆš3*3âˆš3)

D= (2-36)

âˆšD= âˆš(-34)

âˆšD=âˆš34 i

So, roots will be,

R1= (-(âˆš2)+âˆš34 i)/(2*âˆš3) and R2= (-(âˆš2) – âˆš34i)/(2*âˆš3)

Hence, R1= (-âˆš2+âˆš34 i)/(2âˆš3) and R2= (-âˆš2-âˆš34 i)/(2âˆš3).

### Question 21. âˆš2x2 + x + âˆš2 = 0

Solution:

Comparing the equation with,

ax2+bx+c=0

We get, a=âˆš2,b=1,c=âˆš2

Using Discriminant Method,

D= (b2-4ac)

D= ((1)2– 4*âˆš2*âˆš2)

D= (1-8)

âˆšD= âˆš(-7)

âˆšD=âˆš7 i

So, roots will be,

R1= (-(1)+âˆš7 i)/(2*âˆš2) and R2= (-(1) – âˆš7 i)/(2*âˆš2)

Hence, R1= (-1+âˆš7 i)/(2âˆš2) and R2 = (-1-âˆš7 i)/(2âˆš2).

### Question 22. x2 + x + (1/âˆš2) = 0

Solution:

Comparing the equation with,

ax2+bx+c=0

We get, a=1,b=1,c=1/âˆš2

Using Discriminant Method,

D= (b2-4ac)

D= ((1)2 – 4*1*(1/âˆš2))

D= (1-2âˆš2)

âˆšD= âˆš(-(2âˆš2-1))

âˆšD=âˆš(2âˆš2-1) i

So, roots will be,

R1= (-(1)+âˆš(2âˆš2-1) i)/(2) and R2= (-(1) – âˆš(2âˆš2-1) i)/(2)

Hence, R1= (-1+âˆš(2âˆš2-1) i)/(2) and R2= (-1-âˆš(2âˆš2-1) i)/(2).

### Question 23. x2 + (1/âˆš2)x  + 1 = 0

Solution:

Comparing the equation with ,

ax2+bx+c=0

we get ,a=1,b=1/âˆš2,c=1

Using Discriminant Method,

D= (b2-4ac)

D= ( (1/âˆš2)2– 4*1*1)

D= (1/2-4)

âˆšD= âˆš(-7/2)

âˆšD=âˆš(7/2) i

So, roots will be,

R1= (-(1/âˆš2)+âˆš(7/2)i)/2 and R2= (-(1/âˆš2) – âˆš(7/2)i)/2

Hence, R1= (-1+âˆš7i)/(2âˆš2) and R2= (-1-âˆš7i)/(2âˆš2).

### Question 24. âˆš5x2 + x + âˆš5 = 0

Solution:

Comparing the equation with,

ax2+bx+c=0

We get, a=âˆš5,b=1,c=âˆš5

Using Discriminant Method,

D= (b2-4ac)

D= ( (1)2– 4*âˆš5*âˆš5)

D= (1-20)

âˆšD= âˆš(-19)

âˆšD=âˆš19 i

So, roots will be,

R1= (-(1)+âˆš(19)i)/(2*âˆš5) and R2 = (-(1)-âˆš(19) i)/(2*âˆš5)

Hence, R1= (-1+âˆš19i)/(2âˆš5) and R2 = (-1-âˆš19i)/(2âˆš5).

### 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)

âˆšD= âˆš(-7)

âˆšD=âˆš7 i

So, roots will be,

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

Hence, R1= (-1+âˆš7 i)/(-2) and R2= (-1-âˆš7 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)

âˆšD= âˆš(-2)

âˆšD=âˆš2 i

So, roots will be,

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

Hence, R1= (1+i/âˆš2) and R2= (1-i/âˆš2).

### 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)

âˆšD= âˆš(-64)

âˆšD=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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