Tell me about the Vedic theorem

According to Veda's theorem x1+x2=-b a

x -6x+7=0 " where a is 1 and b is -6, so x1+x2=-(-6) 1 is x1+x2=6

According to Veda's theorem x1x2=c a

x -6x+7=0 ", where a is 1 and c is 7, so x1x2=7 1 i.e. x1x2=7

Because (x1+x2) = x1 +x2 +2x1x2 then x1 +x2 =(x1+x2) -2x1x2=22 then apply Veda's theorem.

x1+x2=6 x1x2=7

Substituting (x1+x2) -2x1x2=22

Let the equation have 2 roots, then it is.

x-x1)*(x-x2)=0

x 2-(x1+x2)*x+x1*x2=0 and x 2-6x+7=0 are compared.

x1^2+x2^2=（x1+x2)^2-2x1*x2 = 36-14

Solution: According to the relationship between the root and the coefficient: let the general form of the unary quadratic equation be ax+bx+c=0, then according to the relationship between the root and the coefficient, we can get x1+x2=-b a x1x2=c a Substituting this theorem into your equation and calculating it is x1+x2=6 x1x2=7 (Note:

All of these problems can be done in this way) x of ax is squared. Good luck with your studies!

If you use Vedic theorem, it's simple. Because x1+x2=-b a x1x2=c a

x1 +x2 =(x1+x2) -2x1x2 Then substitute the number and you're done!

1 KnownSatisfies -9=0

So . .Be.

x²-x-9=0

The two of them are dismantled.

7 +3 -66= Eliminate jujube +9 +

And because =9+

Original = 9 + 10 + 10 -3

The original formula = 16 according to Veda's theorem.

=2-m1+mα+α

1+m + hole circle = [1+(m+1)( m+1) 2 ]2 substitution. 9m 2 + 24 m + 16

Example 4 shows that the image of the quadratic function y x2 px q intersects with the x-axis at ( , 0) and (0), and 1 , and verifies that p q 1

97 Sichuan Provincial Junior High School Mathematics Competition Questions).

Proof: From the meaning of the question, we can know that the two roots of the equation x2 px q 0 are , which is obtained by Veda's theorem.

β＝p，αβq．

So p q 1) 1

( 1 ) ( 1 ) 1 1 ( 1 )

1. The answer is 2

x1^2+x2^2=(x1+x2)^2-2x1x2=9-2=7

Denominator: (x1-1)(x2-1)=x1x2-(x1+x2)+1=1-(-3)+1=5

Molecule: x2(x2-1)+x1(x1-1)=x2*x2-x2+x1*x1-x1=x2 2+x1 2-(x2+x1)=7-(-3)=10

So the answer to the whole equation is 2

2. The answer is root number 5

x1-x2=root[(x1-x2) 2] (x1-x2) 2=x1 2+x2 2-2x1x2=(x1 2+x2 2+2x1x2-4x1x2)=(x1+x2) 2-4x1x2=5 So x1-x2=root5

The third question is that there is a formula (a+b)(a 2-ab+b 2)=a 3+b 3

a-b）（a^2+ab+b^2)=a^3-b^3

So this question (x1+x2)(x1 2+x2 2-x1x2)=(-3)(7-1)=-18

Solution: x 3x 10 0

x－5）（x＋2）＝0

x1＝5, x2＝－2

Let the equation be x bx c 0, and its two roots are x1 and x2 respectively.

x1′·x2′＝－2＝c, x1′²＋x2′²＝5

x1′＋x2′）²2x1′x2′＝5

x1′＋x2′）²5＋2x1′x2′＝5＋2×（－2）＝1∴ x1′＋x2′＝±1＝b

x x 2 0 is the required equation. (Actually, this equation has only two imaginary roots).

There is no solution to this new equation. The reasoning is as follows:

The original equation becomes: (x-5)(x+3)=0, and the solution is: x1=5, x2=-3.

Let the two real numbers of the new equation be: y1 and y2

Then because the product of the two real roots and the sum of the squares of the two roots are -3 and 5 respectively, then y1*y2=-3, y1 2+y2 2=5

Then (y1+y2) 2=y1 2+y2 2+2y1*y2=5-6=-1

We all know that the sum of two numbers is squared or greater than or equal to 1, so it is impossible to have the product of two real numbers and the sum of the squares of the two roots of the two numbers being -3 and 5 respectively.

1 KnownMeet -9=0 9=0 soare the two roots of x -x-9=0 = 9 =9+ 7 +3 -66= +9 + 7(9+ )3 -66= +9 +

And because =9+ primitive=9+10 +10 -3 =10( +6 +=1 primitive=16

2 According to Vedder's theorem, +=2-m =1(1+m + 1+m + =[1+(m+1)( m+1) 2 ]2 is substituted to obtain 9m 2+24m+16

Example 4 shows that the image of the quadratic function y x2 px q intersects with the x-axis at ( , 0) and (0), and 1 , and verifies that p q 1

97 Sichuan Provincial Junior High School Mathematics Competition Questions).

Proof: From the meaning of the title, we can know that the two roots of the equation x2 px q 0 are , and p and q are obtained by Veda's theorem

So p q

( 1 ) ( 1 ) 1 1 ( 1 )