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- # include <stdio.h>
; {. n7 z% N" z7 C0 K6 b - # include <math.h>
; f% I; n( F* P# [& R6 B - 1 J2 `- Q# R2 ]) l
- int main(void)) L; o0 [! V( j E) ^' D
- {
7 b& Q% b* R" b9 ]) g' U( w1 { - int a = 1;
" G7 \) u8 E9 J) k. _ - int b = 2;' o- ]& s M* l5 \
- int c = 3;
3 r8 C7 F; M" ?) W - double delta;
2 v6 z9 h0 d! I - double x1;/ Z; U* K3 V. t0 [- w. g* N+ U- W3 @
- double x2;
7 E8 }6 H( t9 n7 T N -
, `# d Q4 x2 D* K - delta = b*b - 4*a*c;
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- if (delta > 0). r4 W' L* }" a- M# D# B, n, H# P
- {
% O& j+ f- s# ]' ]" l; s - x1 = (-b + sqrt(delta)) /(2*a);( l9 l V3 o8 d6 @: |0 x5 \
- x2 = (-b - sqrt(delta)) / (2*a);( \ h4 {9 }1 Y
- printf("该一元二次方程有两个解,x1 = %f,x2 = %f\n", x1, x2);" W8 \$ p: |7 v( p
- }
4 P5 ]1 W" U$ @" l7 h/ ^! q - else if (delta = 0)
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- x1 = (-b) / (2*a);
- y6 q4 K" j/ T9 I& M6 d5 i1 s - x2 = x1;6 z) {% s' A! N' I2 C# H' n0 j
- printf("该一元二次方程有一个唯一解x1 = x2 = %f\n", x1);
' u. U9 S# y% _8 A9 f: |/ X+ I - }
- N, X8 m6 h: `) i - else7 O! R. C5 J: L# I8 C" k
- {" N, U+ H0 `% A. C' e! ]6 p
- printf("无解\n");: C# A9 v4 Q8 E9 X
- }
i) j& P+ M$ H' a& u9 w8 T( R" v% I - return 0;7 M: k- ~1 \% j+ e L4 O
- }& V: f6 N, g# |/ V d9 }
- * l# m: S1 F- g3 {' y% y$ Q; U: p& s$ z
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