I didn t learn Vedic theorem, so I won t

In addition to the Vedic theorem, there are other methods.

Solution: a, b are two unequal real roots of x2 + (2m+3)*x+m 2=0.

a^2+(2m+3)a+m^2=0...

b^2+(2m+3)b+m^2=0...

Get. a^2-b^2+(2m+3)(a-b)=0a+b)(a-b)+(2m+3)(a-b)=0a-b)(a+b+2m+3)=0

A and B are not equal.

a+b+2m+3=0

a+b=-2m-3

1/a + 1/b =(a+b)/ab= -1ab=2m+3

Get. a^2+b^2+(2m+3)(a+b)+2m^2=0a^2+b^2=-(2m+3)(a-b)-2m^2==-(2m+3)(-2m-3)-2m^2=(2m+3)^2-2m^2

and a 2+b 2=(a+b) 2-2ab=(2m+3) 2-2(2m+3).

2m+3)^2-2m^2=(2m+3)^2-2(2m+3)m^2=2m+3

m^2-2m-3=0

The solution yields m=-1 or m=3

It is absolutely impossible not to do without that theorem.

Veda's theorem: the sum of two roots = -b a and the product of two roots = c a So, 1 a + 1 b = -1 can be transformed.

a+b=-ab

So. (2m+3)/1=m^2/1

Solution: m=-3 or 1

This theorem is best remembered. It is very useful when taking the high school entrance examination.

1/a+1/b=a+b/ab

a+b=-2m-3

ab=m^2

2m-3/m^2=-1

m^2+2m-3=0

m=1,-3

Two unequal real roots. m=1

1/a + 1/b = (a+b)/ab

a+b=-(2m+3)

ab=m^2

Substitution can be m=3 or =-1

Two unequal real roots.

m=-1

Vedic theorem:

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

In 1615, the French mathematician François Vedt established the relationship between the roots of equations and the coefficients 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. 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 it by relying on the fundamental theorem of algebra, which was only first substantially demonstrated by Gauss in 1799.

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

Vedic theorem highlights its unique usefulness in finding the symmetry scale of the root, discussing the symbol of the root of the quadratic equation, solving the symmetric equation nucleus, and solving some problems related to the quadratic curve.

The relationship between Vedder's theorem and the discriminant formula of the root of a quadratic equation is even more inseparable.

If ax 2+bx+c=0 has two roots, x1 and x2, then x1+x2= -b a x1*x2=c a

It's very important to argue. In high school, Yuyuan is often used in the vertical stove state, and it is necessary to learn and know it.

Vedic theorem ,,, solution of a quadratic equation.

Let the equation for the ellipse be, and then build a system of equations according to the problem.

Get x1+x2=-2 x1*x2=-5 and classify the sought transformation into the above two forms.

1) x1 2+x1 2 to square.

Original =(x1+x2) 2-2 x1*x2=(-2) 2-2*(-5)=14

2) 1 x1 + 1 x2 pass-through.

Original = (x1+x2) x1*x2=-2 (-5)= 2 5(3)(x1-5)(x2-5) Multiplication of the original formula = x1*x2-5(x1+x2)+25=-5-5*(-2)+25=30

4）x1-x2=± √x1+x2）^2-4 x1*x2)=± √2)^2-4*(-5))=±2√6

Because a is not equal to 0, so x=0 is not the root of the equation cx 2+bx+a=0 x=1 y, bring in the equation to get c+by+ay 2=0 using the known conditions to know that all the roots of the above equation are y=2 and 3, so x=1 y=1 2 and 1 3 are the two roots of the equation cx 2+bx+a=0, and they are also all the roots.

No need for the Great Theorem at all!

ax 2+bx+c=o are two real numbers with roots of 2 and 3

By the Vedic theorem.

2+3=-b/a

2*3=c/a

So b = -5a, c = 6a

cx^2+bx+a=0

6ax^2-5ax+a=0

6x^2-5x+1=0

3x-1)(2x-1)=0

x=1/3,x=1/2

According to the theorem -

2+3=-b/a=5

2*3=c/a=6

Let the roots of cx 2+bx+a=0 be x1, x2x1+x2=-b c=(-b a) (c a)=5 6x1*x2=a c=1 6

Then solve the above two equations to get :

x1=1/2 x2=1/3

Hello, this kind of question is generally common in conic curves, first recommend a question and give you a try for this kind of question.

Some of the answers were revealed, and the answer to the second question was (0,4).

Is it "Veda's theorem" [used in reverse]?

For example, if we already know a+b=-5 and ab=4, then [reversely] Veda's theorem, a and b can be considered as equations.

x 2+5x+4=0. (Of course, the inverse Vedic theorem [does not] have uniqueness - many equations can be corresponded to by conditions.) ）

Similarly, if you know e m+2e n e (m+n)=50

Then we can say that e m, 2e n are the solutions of the equation x 2-15x+50=0.

Method: The coefficient of the quadratic term is the coefficient of the first term in the equation after changing the plus and minus signs of the two numbers, and the product of the two numbers is used as the constant term.