If the ratio of the two roots of Eq. ax 2 bx c 0 is 1 3, the relationship between abc is

From Veda's theorem:

The two real roots of this equation x1+x2=-b a, x1*x2=c a, let x2=3x1

Then 4x1=-b a, 3*(x1) 2=c a is obtained by x1=-b 4a, substituting , b 2 16a 2=c 3a is simplified to: 3b 2=16ac

This is the condition for satisfying the x1 constant abc of x2 = 3 times.

So choose B.

Let one be x and the other be 3x

x+3x=-b/a===> 4x=-b/a , x=-b/4a ===> 4x²=b²/(4a²)===>x²=b²/16a²

x*(3x)=c a ===> x =c 3a, i.e., b 16a =c 3a ===>3b = 16ac to choose b

Let one be x and the other be 3x

x+3x=-b a gives 4x=-b a x=-b (4a)x*x=b*b (16a*a).

x*3*x=c a gives x*x=a (3c) and then b*b (16a*a)=c (3a)3b*b=16ac

So choose B, I hope it can help you.

Let one root be x, then the other root is 3x, which is related to the coefficient by the root.

x+3x=-b a gives 4x=-b a x=-b (4a)x*x=b*b (16a*a).

x*3*x=c a gives x*x=a (3c). b*b (16a*a)=c (3a)3b*b=16ac

Because the equation ax 2+bx+c=0(c≠0) has a root of -1

According to the definition of the equation of the reed socks, x=-1 holds the original equation.

So Biji a-b+c=0

1: When x=1.

1-a+b=0

b+1=a2: when x=3.

9-3a+b=0

Bring a=b+1 into 9-3a+b.

9-3（b+1）=0

9-3b-3=0

3b=-9+3

b=2, so a=3

Substituting 1 into the equation, i.e., 1-a+b=0;Substituting 3 into the equation, i.e., 9-3a+b=0.

Eq. 1-Eq. 2, i.e., 1-a+b-(9-3a+b)=0, simplified to obtain 3a-a=9-1, solution a=4, substitution of Eq. 1, i.e., 1-4+b=0, solution b=3.

then a=4, b=3.

i.e. x1+x2=-1

x1-x2=1

x1=0x2=-1

x1 and x2 are substituted into the original formula 0

When x=0 gives a=c

When x=-1 gets.

a=b, therefore.

A = b = c triangle is an equilateral triangle.

Solution: (1) Let x1=2k, x2=3k

x1+x2=-b/a=5k

x1*x2=c/a=6k^2

(x1+x2) 2 x1*x2).

b^2/ac = 25/6

So 6b 2=25ac

2) Conjecture: (mn)*b 2=(m+n) 2*ac set x1=mk, x2=nk

x1+x2=-b/a=（m+n）k

x1*x2=c/a=mn*k^2

(x1+x2) 2 x1*x2).

b 2 ac = m+n) 2 (mn) so (mn)*b 2=(m+n) 2*ac

Solution: Let the two equations be x1 and x2 respectivelythen 2x1=3x2 (x1, x2 swapping has no effect on the result).

By the meaning of the title, know:

x1+x2=-b/a,x1x2=c/a.

Substituting x1=x2*3 2 into it, you get:

x2*5/2=-b/a...1)

x2)^2*3/2=c/a...2)

After squared (1) and divided by (2), we get:

6*b^2=25*ac.

Note: The key to this problem is to obtain three equations from the relationship between the roots and the coefficients of the unary quadratic equation and the ratio of the two roots to 2:3.

By substituting and eliminating x1 and x2, we can get the relationship between a, b, and c.

Let the two roots be x1 and x2, then there is x1 x2 = 2 3, so x2 = 3x1 2, and because x1 + x2 = -b a, x1x2 = c a, x1 = -2b 5a, x2 = -3b 5a, x1x2 = (-2b 5a) (-3b 5a) = c a, and the solution is 6b 2 = 25ac

Let x1:x2=k.So x1 = 2k, x2 = 3k

Relationship between root and coefficient: x1+x2=-b a; x1*x2=c/a

So -b a=5k; x1*x2=6k 2 is substituted by -b a=5k k b 5a into another equation: 6b 2 25ac

Let the two roots be 2k and 3k respectively, then:

2k+3k=-b/a

2k*3k=c/a

i.e. 5k = -b a(1).

6k^2=c/a(2)

1) Square both sides of the formula.

k 2 = b 2 (25a 2) (3) substitute (3) for (2).

6*b^2/(25a^2)=c/a

6b^2=25ac