第 11 课 RD Sharma 解 – 第 14 章二次方程 – 练习 14.1 |设置 1

求解以下二次方程：

问题 1. x ² + 1 = 0

解决方案：

We can write the given equation as,

x² – i² =0, where i = iota = √(-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.

问题 2。9x ² + 4 = 0

解决方案：

  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.

问题 3. x ² + 2x + 5 = 0

解决方案：

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.

问题 4. 4x ² – 12x + 25 = 0

解决方案：

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). 

问题 5. x ² + x + 1 = 0

解决方案：

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 – ((√3)/2 i)²=0

(x+1/2+ (√3)/2 i)(x+1/2-(√3)/2 i)=0

So, (x+1/2+ (√3)/2 i)=0 and (x+1/2-(√3)/2 i)=0

x=(-1-(√3)i)/2 and x=(-1+(√3)i)/2 

Hence, roots will be x=(-1-(√3))/2 i and x=(-1+(√3))/2

问题 6. 4x ² + 1 = 0

解决方案：

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.

问题 7. x ² – 4x + 7 = 0

解决方案：

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)

√D= √(-12)= 2√3 i

So, roots will be 

R1= (-(-4) + 2√3 i)/2 and R2= (-(-4) – 2√3 i)/2

R1= 2+√3 i and R2= 2-√3i

问题 8. x ² + 2x + 2 = 0

解决方案：

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)

√D = √(-4)

√D= 2i

So, roots will be,

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

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

问题 9. 5x ² – 6x + 2 = 0

解决方案：

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)

√D = √(-4)

√D = 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.

问题 10. 21x ² + 9x + 1 = 0

解决方案：

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)

√D= √(-3)

√D=√3 i

So, roots will be,

R1= (-(9)+ √3 i)/(2*21) and  R2= (-(9) – √3 i)/(2*21)

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

问题 11. x ² – x + 1 = 0

解决方案：

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)

√D= √(-3)

√D=√3 i

So, roots will be,

R1= (-(-1)+ √3 i)/2 and R2= (-(-1) – √3 i)/2

Hence, R1= (1+√3i)/2 and R2= (1-√3i)/2.

问题 12. x ² + x + 1 = 0

解决方案：

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)

√D= √(-3)

√D=√3 i

So, roots will be,

R1= (-(1)+ √3 i)/2 and R2 = (-(1) – √3 i)/2

Hence, R1= (-1+√3i)/2 and R2= (-1-√3i)/2.

问题 13. 17x ² – 8x + 1 = 0

解决方案：

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)

√D= √(-4)

√D=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.