11类NCERT解决方案-第5章复数和二次方程式-练习5.3

求解以下每个方程式：

问题1. x ² + 3 = 0

解决方案：

We have,

x² + 3 = 0 or x² + 0 × x + 3 = 0 —(1)

Discriminant, D = b² – 4ac

from (1) , a = 1 , b = 0 and c = 3

D = (0)² – 4*(1)*(3)

D = -12 

Since , x = $\frac{( -b ± \sqrt{D} }{2a}$

Therefore , 

x = $\frac{( -0 ± \sqrt{-12} }{2(1)}$

x = $\frac{(± 2\sqrt{3}i }{2}$

x = ± √3 i

Hence , the solution of x² + 3 = 0 is ± √3 i

问题2. 2x ² + x + 1 = 0

解决方案：

We have,

2x² + x + 1 = 0 —(1)

Discriminant, D = b² – 4ac

from (1) , a = 2 , b = 1 and c = 1

D = (1)² – 4*(2)*(1)

D = -7

Since , x = $\frac{( -b ± \sqrt{D} }{2a}$

Therefore ,

x = $\frac{( -1 ± \sqrt{7} i}{2(2)}$

x = $\frac{( -1 ± \sqrt{7} i}{4}$

Hence , the solution of  2x² + x + 1 = 0  is $\frac{( -1 ± \sqrt{7} i}{4}$ .

问题3. x ² + 3x + 9 = 0

解决方案：

We have,

x² + 3x + 9 = 0 —(1)

Discriminant, D = b² – 4ac

from (1) , a = 1 , b = 3 and c = 9

D = (3)² – 4*(1)*(9)

D = -27

Since , x = $\frac{( -b ± \sqrt{D} }{2a}$

Therefore ,

x = $\frac{( -3 ± \sqrt{27}i }{2(1)}$

x = $\frac{( -3 ± 3\sqrt{3}i }{2}$

Hence , the solution of  x² + 3x + 1 = 0  is $\frac{( -3 ± 3\sqrt{3}i }{2}$ .

问题4. -x ² + x – 2 = 0

解决方案：

We have,

-x² + x – 2 = 0 —(1)

Discriminant, D = b² – 4ac

from (1) , a = -1 , b = 1 and c = -2

D = (1)² – 4*(-1)*(-2)

D = -7

Since , x = $\frac{ -b ± \sqrt{D} }{2a}$

Therefore ,

x = $\frac{ -1 ± \sqrt{7} i}{2(-1)}$

x = $\frac{ -1 ± \sqrt{7} i}{-2}$

Hence , the solution of  -x² + x – 2= 0  is $\frac{ -1 ± \sqrt{7} i}{-2}$ .

问题5. x ² + 3x + 5 = 0

解决方案：

We have,

x² + 3x + 5 = 0 —(1)

Discriminant, D = b² – 4ac

from (1) , a = 1 , b = 3 and c = 5

D = (3)² – 4*(1)*(5)

D = -11

Since , x = $\frac{ -b ± \sqrt{D} }{2a}$

Therefore ,

x = $\frac{ 3 ± \sqrt{11}i }{2(1)}$

x = $\frac{ 3 ± \sqrt{11}i }{2}$

Hence , the solution of  -x² + x – 2= 0  is $\frac{ 3 ± \sqrt{11}i }{2}$ .

问题6. x ² – x + 2 = 0

解决方案：

We have,

x² – x + 2 = 0 —(1)

Discriminant, D = b² – 4ac

from (1) , a = 1 , b = -1 and c = 2

D = (-1)² – 4*(1)*(2)

D =  -7

Since , x = $\frac{ -b ± \sqrt{D} }{2a}$

Therefore ,

x = $\frac{ -(-1) ± \sqrt{-7} }{2(1)}$

x = $\frac{ 1 ± \sqrt{7} }{2}$

Hence , the solution of  -x² + x – 2= 0  is $\frac{ 1 ± \sqrt{7} }{2}$ .

问题^7.√2×2 + X +√2= 0

解决方案：

We have,

√2x² + x + √2 = 0 —(1)

Discriminant, D = b² – 4ac

from (1) , a = √2 , b = 1 and c = √2

D = (1)² – 4*(√2)*(√2)

D =  -7

Since , x = $\frac{ -b ± \sqrt{D} }{2a}$

Therefore ,

x = $\frac{ -1 ± \sqrt{7} i}{ 2(\sqrt{2})}$

Hence , the solution of  -x² + x – 2= 0  is $\frac{ -1 ± \sqrt{7} i}{ 2\sqrt{2}}$ .

问题^8.√3×2 – √2x+3√3= 0

解决方案：

We have,

√3x² – √2x + 3√3 = 0 —(1)

Discriminant, D = b² – 4ac

from (1) , a = √3 , b = -√2 and c = 3√3

D = (-√2)² – 4*(√3)*(3√3)

D =  -34

Since , x = $\frac{ -b ± \sqrt{D} }{2a}$

Therefore ,

x = $\frac{-(-\sqrt{2} ± \sqrt{34}i)}{ 2(\sqrt{3})}$

x =  $\frac{\sqrt{2} ± \sqrt{34}i}{ 2(\sqrt{3})}$

Hence , the solution of  -x² + x – 2= 0  is $\frac{\sqrt{2} ± \sqrt{34}i}{ 2(\sqrt{3})}$.

问题9. x ² + x + $\frac{1}{\sqrt{2}}$ = 0

解决方案：

We have,

x² + x + $\frac{1}{\sqrt{2}}$ = 0  or  √2x² + √2x + 1 = 0 —(1)

Discriminant, D = b² – 4ac

from (1) , a = √2 , b = √2 and c = 1

D = (√2)² – 4*(√2)*(1)

D =  2 – 4√2 = 2 ( 1 – 2√2 )

Since , x = $\frac{ -b ± \sqrt{D} }{2a}$

Therefore ,

x = $\frac{ -\sqrt{2} ± \sqrt{2(1-2\sqrt{2}})}{2\sqrt{2}}$

x = $\frac{ -1 ± \sqrt{2(\sqrt{2}-1)}i}{2}$

x = $\frac{ -1 ± \sqrt{(2\sqrt{2}-1}i}{2}$

Hence , the solution of  -x² + x – 2= 0  is $\frac{ -1 ± \sqrt{(2\sqrt{2}-1}i}{2}$.

问题10.x ² + $\frac{x }{ \sqrt{2} }$ +1 = 0

解决方案：

We have,

x² + $\frac{x }{ \sqrt{2} }$ + 1 = 0  or √2x² + x + √2 = 0 —(1)

Discriminant, D = b² – 4ac

from (1) , a = √2 , b = 1 and c = √2

D = (1)² – 4*(√2)*(√2)

D =  -7

Since , x = $\frac{ -b ± \sqrt{D} }{2a}$

Therefore ,

x = $\frac{ -1 ± \sqrt{7}i }{2 (\sqrt{2})}$

x = $\frac{ -1 ± \sqrt{7}i }{2\sqrt{2}}$

Hence , the solution of  x² + $\frac{x }{ \sqrt{2} }$ + 1 = 0  is $\frac{ -1 ± \sqrt{7}i }{2\sqrt{2}}$.