On the question of Vedic theorem, how is Vedic theorem understood?

Solution: Let the two integer roots of the equation be x1 and x2, you may wish to set x1 x2 So the original equation can be written as: (x-x1)(x-x2)=0 to get x 2-(x1+x2)x+x1x2=0

Compared with the original equation, we can see that p=-(x1+x2); q=x1x2 so there is. x1＋x2＝－p，x1x2＝q．

This is the **, essence of the Vedic theorem!

Got it? Good luck with your studies!

Vedica's theorem states that for ax 2+bx+c=0(a≠0,b 2-4ac 0), if x1 and x2 are the two roots of the equation, then there is x1+x2=-b a,x1*x2=c a.

So"x1＋x2＝－p "That's right.

For ax 2+bx+c=0(a≠0,b 2-4ac 0), if x1 and x2 are the two roots of the equation, then x1=(-b+(b 2+4ac) (1 2)) 2a,x2=(-b-(b 2+4ac) (1 2)) 2a, then x1+x2=-b a; x1*x2=c/a;This is also known as the Vedic theorem. At the same time, we can also see the reason for a≠0, b2-4ac 0.

As a quadratic equation, the sum of the two roots of ax2+bx+c=0 is -b a, and the product of the two roots is c a

Da theorem: Let a quadratic equation be a one.

, the two x and x have the following relationship:

Vedder's theorem explains the relationship between roots and coefficients in a quadratic equation. Suspicion.

In 1615, the French mathematician François Veda established the relationship between the root of the equation and the Lu's coefficient in his work "On the Identification and Revision of Equations", and proposed this theorem.

Because Veda first developed this relationship between the roots and the coefficients of modern number equations, people call this relationship Veda's theorem.

Veda's theorem explains the relationship between roots and coefficients in a univariate nth order equation.

Here we will mainly talk about the relationship between the two roots of a one-dimensional quadratic equation.

The unary quadratic equation ax 2+bx+c=0 (a≠0 and b 2-4ac 0), two. x1x2

There is the following relationship:

x1+x2=-b/a;

x1*x2=c/a.

The unary quadratic equation ax 2+bx+c=0

a≠0 and =b 2-4ac 0).

Let the two roots be x1 and x2

then x1+x2=

b/ax1*x2=c/a

Use the Vedic theorem to judge the root of the equation.

If b 2-4ac>0

Then the equation has two unequal real roots.

If b 2-4ac>0

then the equation has two equal real roots.

If b 2-4ac 0 then the equation has a real root.

If b 2-4ac "Stuffy limb 0

then the equation has no real solution.

The Vedic theorem can also be used in higher order equations. In general, for a unary n-order equation aix i=0

Its roots are denoted as x1, x2....,xn

We have. Liquid xi=(-1) 1*a(n-1) a(n).

xixj=(-1)^2*a(n-2)/a(n)

xi=(-1)^n*a(0)/a(n)

where is the sum and is the product.

If a quadratic equation.

The root in the complex set is, then.

From the fundamental theorem of algebra can be deduced: any unary. Equations of the nth order.

There must be roots in the plural. Thus, the left end of this equation can be decomposed into the product of a factor in the range of complex numbers:

where are the roots of the equation. The coefficient of comparison between the two ends is known as the Vedic theorem.

The French mathematician Veda was the first to discover this relationship between the roots and coefficients of modern number equations, so people call this relationship Vedt's theorem. History is interesting, Veda arrived at this theorem in the 16th century, and proved that this theorem depends on the fundamental theorem of algebra, which was only made by Gauss in 1799.

The Vedic theorem has a wide range of applications in equation theory.

x1-x2) is (b 2-4ac under the root number) (absolute value of a).

It is the relationship between the root and the coefficient of a quadratic equation.

Suppose there are two real roots x1 in the equation ax +bx+c=0, and x2 then x1+x2=-b a

x1*x2=c/a

Veda's theorem explains the relationship between roots and coefficients in a univariate nth order equation.

Here we talk about the relationship between the two roots of a quadratic equation.

In the unary quadratic equation ax 2+bx+c=0 δ 0, the two x1 and x2 have the following relationship: x1+ x2=-b a, x1·x2=c a

For a quadratic equation ax +bx+c=0(a≠0).

When there is a solution to this equation, let the two solutions be x1 and x2

then x1+x2=-b a,,x1x2=c a

Yes, it is written like this, for the quadratic equation ax +bx+c=0, the product of its two roots x1 and x2 is c a, and the sum is -b a.

Solution: 1. Because another root is 1, substituting the formula can obtain m=16, then the formula is 3x -19x+16=0, then it can be simplified as.

3x-16)(x-1)=0, then the second root is (16 3);

2. Assuming that the two roots are a and b, then there is a +b = 7, and according to Veda's theorem, there can be a + b = m, ab = 2m-1, and a +b = (a + b) -2ab = m -2(2m-1) = 7, and the simplified m -4m-5 = (m-5)(m+1) = 0, then there is m = 5 or -1, but because when m = 5 is, the equation is meaningless, that is, there are no two real roots, so the final value of m is (-1).

Dear, do you understand???

1.Knowing that one of the equations 3x -19x+m=0 is 1, find its other root and the value of m.

From Veda's theorem:

The sum of the two = 19 3

So, another root = 19 3 - 1 = 16 3

The product of two roots = m 3

1*16/3=m/3

m=162.Knowing that the sum of squares of the two real roots of the equation x -mx + 2m - 1 = 0 about x is 7, find the value of m.

Let the two be x1 and x2

From Veda's theorem:

x1+x2=m

x1x2=2m-1

x1²+x2²=7

x1+x2)²-2x1x2=7

m²-4m+2=7

m²-4m-5=0

m-5)(m+1)=0

m=5 or m=-1