Given that the lengths of the two sides of a straight triangle are the two roots of the equation x-14x + 48 = 0, then the third side of the triangle is []

Given that the lengths of the two sides of a straight triangle are the two roots of the equation x-14x + 48 = 0, then the third side of the triangle is []

10 or 2 root sign 7 first find two solutions, which are 6 and 8 respectively, when they are two right angle sides, the third is 10. It can also be that 8 is a hypotenuse. The third side is 8 minus 6, and the root sign is 2 root sign 7

It is known that in the RT triangle, where both sides are exactly the two roots of the equation xx-14x + 48 = 0, then the length of the hypotenuse of the right triangle is

The roots of the equation are 6 and 8. According to Pythagorean theorem, the length of the hypotenuse is 10

It is known that the quadratic equation of one variable (T-5) x ^ 2-4x-1 = 0 has two real roots, then the value range of T

Because it's a quadratic equation of one variable
So the coefficient of quadratic term is not 0
So T-5 ≠ 0, that is t ≠ 5
Furthermore, (T-5) x ^ 2-4x-1 = 0 has two real roots
Then discriminant
△=b²-4ac
=4²+4(t-5)≥0
16+4t-20≥0
4t≥4
t≥1
In conclusion, the value range of T is t ≥ 1 and t ≠ 5

If the quadratic equation AX ^ 2 + BX + C = 0 and the two roots are X1 = - 4 and X2 = 2, then the quadratic function y = ax ^ 2 + BX + C is the symmetry axis of the image

The axis of symmetry is x = (- 4 + 2) / 2 = - 1

If the two roots of the equation AX & # 178; + BX + C = 0 (a ≠ 0) are x1, X2, then ax & # 178; + BX + C=
Completion

There is something wrong with this topic

Solving equation AX & # 178; + BX + C = 0 (a ≠ 0) by collocation method

Collocation method: use collocation method to solve the equation AX ^ 2 + BX + C = 0 (a ≠ 0)
First move the constant C to the right of the equation: ax ^ 2 + BX = - C
The quadratic coefficient is changed to 1: x ^ 2 + B / AX = - C / A
On both sides of the equation, add the square of half of the coefficient of the first term: x ^ 2 + B / ax + (B / 2a) ^ 2 = - C / A + (B / 2a) ^ 2;
The left side of the equation becomes a complete square formula: (x + B / 2a) 2 = - C / A + (B / 2a) & #;
When B & # 178; - 4ac ≥ 0, x + B / 2A = ± √ (- C / a) + (B / 2a) &# 178;
Ψ x = {- B ± [√ (B & # 178; - 4ac)]} / 2A (this is the root formula)

Using C + + 6.0 to solve the equation AX & # 178; + BX + C = 0
According to the knowledge of algebra, there should be the following possibilities:
(1) A = 0, not a quadratic equation, but a linear equation
(2) B & # 178; - 4ac = 0, has two equal real roots
(3) B & # 178; - 4ac > 0, has two unequal real roots
（4）b²—4ac0)
\x05printf("x1=%f,x2=%f",x1=((-b+sqrt(b*b-4*a*c))/2*a),x2=((-b-sqrt(b*b-4*a*c))/2*a));
\x05else if (b*b-4*a*c==0)
\x05printf("x1=%f",x1=(-b)/2*a);
\X05else printf ("has two conjugated complex roots'";
}
I always run it wrong,

#include
#include
void main ()
{
\x05double a,b,c,x1,x2;
\x05printf("please enter a,b,c:");
\x05scanf("%lf,%lf,%lf",&a,&b,&c);
if (a==0)
\x05 printf("x1=%f",x1=(-c)/b);
\x05else if (b*b-4*a*c>0)
\x05printf("x1=%f,x2=%f",x1=((-b+sqrt(b*b-4*a*c))/2*a),x2=((-b-sqrt(b*b-4*a*c))/2*a));
\x05else if (b*b-4*a*c==0)
\x05printf("x1=%f",x1=(-b)/2*a);
\X05else printf ("has two conjugated complex roots'";
}
That is, a = = 0, you are running once, ask again if you have any questions

A solution equation, X & # 178; + AX = BX (a, B are known numbers)

x²+ax=bx
x(x+a-b)=0
X = 0 or x + A-B = 0
X = 0 or x = B-A

What is the difference between the simplest form AX = B {A is not equal to o} and the general form ax + B = O {A is not equal to o}?

The solutions of x = A / B; X = - A / b are opposite to each other

If ax + B = 0 is a linear equation of one variable with respect to x, then a satisfies

a≠0