第 11 课 RD Sharma 解 – 第 14 章二次方程 – 练习 14.1 |设置 2
问题 14. 27x 2 – 10x + 1 = 0
解决方案:
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.
问题 15. 17x 2 + 28x + 12 = 0
解决方案:
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.
问题 16. 21x 2 – 28x + 10 = 0
解决方案:
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.
问题 17. 8x 2 – 9x + 3 = 0
解决方案:
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.
问题 18. 13x 2 + 7x + 1 = 0
解决方案:
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.
问题 19. 2x 2 + x + 1 = 0
解决方案:
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.
问题 20. √3x 2 – √2x + 3√3 = 0
解决方案:
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).
问题 21. √2x 2 + x + √2 = 0
解决方案:
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).
问题 22. x 2 + x + (1/√2) = 0
解决方案:
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).
问题 23. x 2 + (1/√2)x + 1 = 0
解决方案:
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).
问题 24. √5x 2 + x + √5 = 0
解决方案:
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).
问题 25. -x 2 + x – 2 = 0
解决方案:
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).
问题 26. x 2 – 2x + 3/2 = 0
解决方案:
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).
问题 27。3x 2 – 4x + 20/3 = 0
解决方案:
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.