通过分解求解以下二次方程
问题1.(x – 4)(x + 2)= 0
解决方案:
We have equation,
(x – 4) (x + 2) = 0
Implies that either x – 4 = 0 or x + 2 = 0
Therefore, roots of the equation are 4 or -2.
问题2.(2x + 3)(3x – 7)= 0
解决方案:
We have equation,
(2x + 3) (3x – 7) = 0
Implies that either 2x + 3 = 0 or 3x – 7 = 0
Therefore, roots of the equation are -3/2 or 7/3.
问题3 3×2 – 14× – 5 = 0
解决方案:
We have equation,
3x2 – 14x – 5 = 0
We can factorize this equation as
3x2 – 15x + x – 5 = 0
3x (x – 5) -1 (x – 5) = 0
(3x – 1) (x – 5) = 0
Therefore, roots of the equation are -1/3 or 5.
问题4. 9x 2 – 3x – 2 = 0
解决方案:
We have equation,
9x2 – 3x – 2 = 0
We can factorize this equation as
9x2 – 6x + 3x – 2 = 0
3x(3x – 2) + 1(3x – 2) = 0
(3x – 2) (3x + 1) = 0
Therefore, roots of the equation are 2/3 or -1/3.
问题5.(1 /(x – 1))–(1 /(x + 5))= 6/7,x≠1,-5。
解决方案:
We have equation,
(1/(x – 1)) – (1/(x + 5)) = 6/7
We can rewrite this equation as
(x + 5 – x + 1)/((x – 1) (x + 5)) = 6 / 7
1/((x – 1) (x + 5)) = 1/7, or
(x – 1) ( x + 5) = 7
x2 + 4x – 12 = 0
We can factorize this equation as
x2 + 6x – 2x – 12 = 0
x (x + 6) – 2 (x + 6) = 0
(x – 2) (x + 6) = 0
Therefore, roots of the equation are 2 or -6.
问题6. 6x 2 + 11x + 3 = 0
解决方案:
We have equation,
6x2 + 11x + 3 = 0
We can factorize this equation as
6x2 + 9x + 2x + 3 = 0
3x (2x + 3) +1 (2x + 3) = 0
(2x + 3) (3x + 1) = 0
Therefore, roots of the equation are -3/2 or -1/3.
问题7. 5X 2 – 3× – 2 = 0
解决方案:
We have equation,
5x2 – 3x – 2 = 0
We can factorize this equation as
5x2 – 5x + 2x – 2 = 0
5x (x – 1) + 2(x – 1) = 0
(x – 1) (5x + 2) = 0
Therefore, roots of the equation are -2/5 or -1.
问题8. 48x 2 – 13x – 1 = 0。
解决方案:
We have equation,
48x2 – 13x – 1 = 0
We can factorize this equation as
48x2 – 16x + 3x – 1 = 0
16x (3x – 1) + 1 (3x – 1) = 0
(3x – 1) (16x + 1) = 0
Therefore, roots of the equation are -1/16 or 1/3.
问题9. 3x 2 = -11x – 10。
解决方案:
We have equation,
3x2 + 11x +10 = 0
We can factorize this equation as
3x2 + 6x + 5x + 10 = 0
3x (x + 2) + 5 (x + 2) = 0
(3x + 5) (x + 2) = 0
Therefore, roots of the equation are -2 or -5 / 3.
问题10.25x(x + 1)= -4。
解决方案:
We have equation,
25x2 + 25x + 4 = 0
We can factorize this equation as
25x2 + 20x + 5x + 4 = 0
5x (5x + 4) + 1 (5x + 4) = 0
(5x + 4) (5x + 1) = 0
Therefore, roots of the equation are -1/5 or -4/5.
问题11。16x–(10 / x)= 27
解决方案:
We have equation 16x2 -27x -10 = 0
We can factorize this equation as:
16x2 – 32x + 5x – 10 = 0
16x (x – 2) + 5 (x – 2) = 0
(16x + 5) (x – 2) = 0
Therefore, the roots of equations are 2 or -5/16.
问题12.(1 / x)–(1 /(x – 2))= 3,x≠0,2
解决方案:
We have equation 3x2 – 6x + 2 = 0
Here, a = 3, b = -6 and c = 2
Since,
Discriminant (D) = b2 – 4ac and x = (-b ± √D)/2a
Therefore,
D = 36 – 24 = 12, and
x = (-(-6) ± √12)/6
x = (6 ± 2√3)/6
x = (3 ± √3)/3
Therefore, the roots of equations are (3 + √3)/3 or (3 – √3)/3.
问题13. x –(1 / x)= 3,x≠0
解决方案:
We have equation x2 – 3x -1 = 0
Here, a = 1, b = -3 and c = -1
Since,
Discriminant (D) = b2 – 4ac and x = (-b ± √D)/2a
Therefore,
D = 9 + 4 = 13, and
x = (-(-3) ± √13)/2
x = (3 ± √13)/2
Therefore, the roots of equations are (3 + √13)/2 or (3 – √13)/2.
问题14.(1 /(x + 4))–(1 /(x – 7))= 11/30,x≠4,7
解决方案:
We have equation,
(1/(x + 4)) – (1/(x – 7)) = 11/30
(x – 7 – x – 4)/((x + 4) (x – 7)) = 11/30
-11/((x – 4) (x – 7)) = 11/30
-1/((x – 4) (x – 7)) = 1/30
x2 – 3x + 2 = 0
We can factorize this equation as:
x2 – 2x – x + 2 = 0
x (x – 2) – 1 (x – 2) = 0
(x -1) (x – 2) = 0
Therefore, the roots of equations are 2 or 1.
问题15((1 /(x – 3))+(2 /(x – 2))= 8 / x,x≠0,2,3
解决方案:
We have equation,
(1/(x – 3)) + (2/(x – 2)) = 8/x
(x – 2 + 2x – 6)/((x – 3) (x – 2)) = 8/x
(3x – 8)/((x – 3) (x – 2)) = 8/x
8 ((x – 3) (x – 2)) = x(3x – 8)
5x2 – 32x + 48 = 0
We can factorize this equation as:
5x2 – 20x -12x + 48 = 0
5x (x – 4) -12(x – 4) = 0
(5x – 12) (x – 4) = 0
Therefore, the roots of equations are 12/5 or 4.
问题16. a 2 x 2 – 3abx + 2b 2 = 0
解决方案:
We have equation,
a2x2 – 3abx + 2b2 = 0
We can factorize this equation as:
a2x2 – 2abx – abx + 2b2 = 0
ax (ax – 2b) – b (ax – 2b) = 0
(ax – b) (ax – 2b) = 0
Therefore, the roots of the equation are b/a or 2b/a.
问题17. 9x 2 – 6b 2 x –(a 4 – b 4 )= 0
解决方案:
We have equation,
9x2 – 6b2x – (a4 – b4) = 0
We can factorize this equation as:
(3x)2 – 6b2x + (b2)2 – a4 = 0
(3x – b2)2 – (a2)2 = 0
(3x – b2 + a2) (3x – b2 – a2) = 0
Therefore, the roots of the equation are (b2 – a2)/3 or (b2 + a2)/3.
问题18。4x 2 + 4bx –(a 2 – b 2 )= 0
解决方案:
We have equation,
4x2 + 4bx – (a2 – b2) = 0
Dividing by 4
x2 + bx – ((a2 – b2)/4) = 0
x2 + bx – (((a – b)/2) ((a + b)/2)) = 0
We can write b as :
b = ((a + b)/2) – ((a – b)/2)
Therefore,
x2 + ((a + b)/2) – ((a – b)/2) x – (((a – b)/2) ((a + b)/2)) = 0
x (x + ((a + b)/2)) – ((a – b)/2) (x + (a + b)/2) = 0
(x + ((a + b)/2)) (x – ((a – b)/2)) = 0
Therefore, the roots of the equation are (a – b)/2 or (-a – b)/2.
问题19. ax 2 +(4a 2 – 3b)x – 12ab = 0。
解决方案:
We have equation,
ax2 + 4a2x – 3bx – 12ab = 0
ax (x + 4a) – 3b (x – 4a) = 0
(ax – 3b) (x + 4a) = 0
Therefore, the roots of the equation are 3b/a or -4a.
问题20. 2x 2 + ax – a 2 = 0。
解决方案:
We have equation,
2x2 + ax – a2 = 0
2x2 + 2ax – ax – a2 = 0
2x (x + a) – a (x + a) = 0
(2x – a) (x + a) = 0
Therefore, the roots of the equation are a/2 or -a.