# Class 11 RD Sharma Solutions – Chapter 14 Quadratic Equations – Exercise 14.1 | Set 1

### Question 1.x2 + 1 = 0

Solution:

We can write the given equation as,

x2 – i2 =0, where i = iota = âˆš(-1)

Now factorizing above equation,

(x + i)(x – i) = 0

So, x + i = 0 and x – i = 0

x = -i and x = +i

Hence, roots will be +i and -i.

### Question 2. 9x2 + 4 = 0

Solution:

We can write the given equation as,

9x2 – 4(i2) = 0

(3x)2 – (2i)2 = 0

(3x – 2i)(3x + 2i) = 0

So, 3x – 2i = 0 and 3x + 2i = 0

x = 2i/3 and x = -2i/3

Hence, roots will be 2i/3 and -2i/3.

### Question 3. x2 + 2x + 5 = 0

Solution:

We can write the given equation as,

(x2 + 2x + 1) + 4 = 0

(x2 + 2x + 1) – 4(i2) = 0

(x + 1)2 – (2i)2 = 0

(x + 1 – 2i)(x + 1 – 2i) = 0

So, (x + 1 – 2i) = 0 and (x + 1 – 2i) = 0

x = -1 + 2i and x = -1 – 2i

Hence, roots will be -1 + 2i and -1 – 2i.

### Question 4. 4x2 – 12x + 25 = 0

Solution:

We can write the above equation as,

4x2-12x+9+16=0

(4x2 -12x +9) – 16(i2)=0

(2x-3)2 – (4i)2=0

(2x-3+4i)(2x-3-4i)=0

So, (2x-3+4i)=0 and (2x-3-4i)=0

x=(3-4i)/2 and x=(3+4i)/2

Hence, roots will be (3/2-2i) and (3/2+2i).

### Question 5. x2 + x + 1 = 0

Solution:

We can write the above equation as,

x2+x+(1/4)+(3/4)=0

(x+1/2)2  – (3/4)(i2)=0

(x+1/2)2 – ((âˆš3)/2 i)2=0

(x+1/2+ (âˆš3)/2 i)(x+1/2-(âˆš3)/2 i)=0

So, (x+1/2+ (âˆš3)/2 i)=0 and (x+1/2-(âˆš3)/2 i)=0

x=(-1-(âˆš3)i)/2 and x=(-1+(âˆš3)i)/2

Hence, roots will be x=(-1-(âˆš3))/2 i and x=(-1+(âˆš3))/2

### Question 6. 4x2 + 1 = 0

Solution:

We can write the above equation as,

4x2-1(i2) = 0

(2x)2-(i) 2=0

(2x-i)(2x+i)=0

So, (2x-i)=0 and (2x+i)=0

x=i/2 and x= -i/2

Hence, roots will be x=-i/2 and x=i/2.

### Question 7. x2 – 4x + 7 = 0

Solution:

Comparing the equation with,

ax2+bx+c=0

We get, a=1,b=-4,c=7

Using Discriminant Method,

D= (b2-4ac)

D= ((-4)2 – 4*1*7)

D= (16 -28)

âˆšD= âˆš(-12)= 2âˆš3 i

So, roots will be

R1= (-(-4) + 2âˆš3 i)/2 and R2= (-(-4) – 2âˆš3 i)/2

R1= 2+âˆš3 i and R2= 2-âˆš3i

### Question 8. x2 + 2x + 2 = 0

Solution:

Comparing the equation with,

ax2+bx+c=0

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

Using Discriminant Method,

D= (b2-4ac)

D= ((2)2 – 4*1*2)

D= (4 – 8)

âˆšD = âˆš(-4)

âˆšD= 2i

So, roots will be,

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

Hence, R1= -1+i and R2=-1-i.

### Question 9. 5x2 – 6x + 2 = 0

Solution:

Comparing the equation with,

ax2+bx+c=0

We get, a=5,b=-6,c=2

Using Discriminant Method,

D = (b2-4ac)

D = ((-6)2 – 4*5*2)

D  = (36- 40)

âˆšD = âˆš(-4)

âˆšD = 2i

So, roots will be,

R1= (-(-6) + (2i))/(2*5) and  R2= (-(-6) – (2i) )/(2*5)

Hence, R1= (3+i)/5 and R2=(3-i)/5.

### Question 10. 21x2 + 9x + 1 = 0

Solution:

Comparing the equation with ,

ax2+bx+c=0

We get, a=21,b=9,c=1

Using Discriminant Method,

D= (b2-4ac)

D= ((9)2– 4*21*1)

D= (81- 84)

âˆšD= âˆš(-3)

âˆšD=âˆš3 i

So, roots will be,

R1= (-(9)+ âˆš3 i)/(2*21) and  R2= (-(9) – âˆš3 i)/(2*21)

Hence, R1= -3/14+âˆš3i/42 and R2= -3/14-âˆš3i/42.

### Question 11. x2 – x + 1 = 0

Solution:

Comparing the equation with,

ax2+bx+c=0

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

Using Discriminant Method,

D= (b2-4ac)

D= ((-1)2– 4*1*1)

D= (1- 4)

âˆšD= âˆš(-3)

âˆšD=âˆš3 i

So, roots will be,

R1= (-(-1)+ âˆš3 i)/2 and R2= (-(-1) – âˆš3 i)/2

Hence, R1= (1+âˆš3i)/2 and R2= (1-âˆš3i)/2.

### Question 12. x2 + x + 1 = 0

Solution:

Comparing the equation with,

ax2+bx+c=0

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

Using Discriminant Method,

D= (b2-4ac)

D= ((-1)2– 4*1*1)

D= (1- 4)

âˆšD= âˆš(-3)

âˆšD=âˆš3 i

So, roots will be,

R1= (-(1)+ âˆš3 i)/2 and R2 = (-(1) – âˆš3 i)/2

Hence, R1= (-1+âˆš3i)/2 and R2= (-1-âˆš3i)/2.

### Question 13. 17x2 – 8x + 1 = 0

Solution:

Comparing the equation with,

ax2+bx+c=0

We get, a=17,b=-8,c=1

Using Discriminant Method,

D= (b2-4ac)

D= ((-8)2– 4*17*1)

D= (64- 68)

âˆšD= âˆš(-4)

âˆšD=2i

So, roots will be,

R1= (-(-8)+ 2i)/(2*17) and  R2= (-(-8) – 2i)/(2*17)

Hence, R1= (4+i)/17 and R2= (4-i)/17.

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