Requirements: keyboard input coefficient a, B, C value, solve the root of quadratic equation

Requirements: keyboard input coefficient a, B, C value, solve the root of quadratic equation

#include#includeint main(){double a,b,c,disc,p,q,x1,x2;scanf("%lf%lf%lf",&a,&b,&c);disc=b*b-4*a*c;if(a != 0){if(disc0)printf("%f%f\n",x1,x2);else printf("%f\n",x1);}}else if(b == 0){if(c == 0)printf("...

To solve the univariate quadratic equation (A & sup2; - B & sup2;) x & sup2; + AB = (A & sup2; + B & sup2;) x

In the quadratic equation AX2 + BX + C = 0, if a + B + C = 0, then one root of the equation must be______ If A-B + C = 0, then one root of the equation must be______ ．

Substituting x = 1 into the original equation, we can get a + B + C = 0, and the root of the equation must be 1; substituting x = - 1 into the original equation, we can get A-B + C = 0, and the root of the equation must be - 1

Please write one root as 1 and the other root satisfies - 2

The simplest form of quadratic equation with one variable is ~ ~ because one root is 1, and the other is optional, if - 1.5
We can get (x-1) (x + 1.5) = 0

The solution of quadratic equation with one variable, MX-1 = 2x, is an integral real number

MX-1 = 2x, mx-2x = 1, (m-2) x = 1, x = 1 / m-2, when m = 3, x = 1

If the two solutions of the quadratic equation AX ^ 2 + BX + C = 0 are 1 and - 1, then the relationship between a and C is

ax^2+bx+c=0
Two solutions are 1 and - 1
consider
a(x-1)(x+1)
= a(x^2-1)
= ax^2 - a
=>
b =0,c = -a
ie
a = -c

As shown in the figure, in △ ABC, ch is the bisector of ∠ ACD, BH is the bisector of ∠ ABC, ∠ a = 58 ° and the degree of ∠ h is calculated

BH and Ca intersect at point O to form two opposing triangles
Therefore, H + 2 = 1 + a
In the triangle BCH, ∠ 2 is the outer angle, so ∠ 2 = 1 + H
Two equations are subtracted
∠H=∠1+∠A-∠1-∠H
2∠H=∠A
∠H=58/2=29°

The quadratic equation AX ^ 2 + BX + C = 0, if A-B + C = 0, then one of its solutions is

This is a quadratic function. When x equals - 1, it's the equation behind you. So x equals - 1. I just graduated from grade 3. We review it all day`

In △ ABC, if SINB (1 + COSA) ≥ (2-cosb) Sina is satisfied, then the value range of a is______ ．

In △ ABC, ∵ SINB (1 + COSA) ≥ (2-cosb) Sina, SINB + sinbcosa + cosbsina ≥ 2sina, that is, SINB + sin (a + b) ≥ 2sina, ∵ SINB + sinc ≥ 2sina, then B + C ≥ a can be obtained from the sine theorem, so edge a is not the largest edge, so a is an acute angle 2sinb + c2cosb − C2 ≥ 2sina, ∧ Sina ≤ 2sinb + C2 = 2Sin (90 ° - A2), ∧ a ≤ 90 ° - A2, ∧ 0 < a ≤ 60 °, so the answer is: (0, 60 °]

Is it known that the quadratic equation AX2 + BX + C = 0 has a root greater than 2 and a root less than 2?

Method 1:
Let your equation be the function f (x)
The intersection of F (x) and X axis is one greater than 2 and one less than 2,
So there are two situations
When a > 0, the opening is upward, then f (2)