1. By solving the equation, we can see that the two symbols of equation 2x & # 178; + 3x-7 = 0 are a. same sign B. different sign C. both of them are positive D. uncertain 1. The two symbols of equation 2x & # 178; + 3x-7 = 0 are () A. The same sign B. different sign C. both are positive D. uncertain 2. If √ 2-1 is a root of X & # 178; - MX-1 = 0, then M is equal to () A.-2 B.-3 C.1 D.2 3. In the following equation, the two real roots of equation () are X1 and X2 respectively, and satisfy X1 + x2 = 3 A.x²+3x+4=0 B.x²+3x-4=0 C.x²-3x+4=0 D.x²-3x-4=0 4. If - 3 is a root of the equation x & # 178; + KX - (K & # 178; - 1) = 0 (k > 0), then the other root is () A. 3 / 2 B.1 C. - 1 D. - 5 / 2

1. By solving the equation, we can see that the two symbols of equation 2x & # 178; + 3x-7 = 0 are a. same sign B. different sign C. both of them are positive D. uncertain 1. The two symbols of equation 2x & # 178; + 3x-7 = 0 are () A. The same sign B. different sign C. both are positive D. uncertain 2. If √ 2-1 is a root of X & # 178; - MX-1 = 0, then M is equal to () A.-2 B.-3 C.1 D.2 3. In the following equation, the two real roots of equation () are X1 and X2 respectively, and satisfy X1 + x2 = 3 A.x²+3x+4=0 B.x²+3x-4=0 C.x²-3x+4=0 D.x²-3x-4=0 4. If - 3 is a root of the equation x & # 178; + KX - (K & # 178; - 1) = 0 (k > 0), then the other root is () A. 3 / 2 B.1 C. - 1 D. - 5 / 2

Two are set as x1, x2
X1*X2=(-7)/2

Solving quadratic equation AX ^ + BX = C = 0 (a is not equal to 0) with collocation method

X & # 178; + BX / a = - C / ax & # 178; + BX / A + B & # 178; / 4A & # 178; = B & # 178; / 4A & # 178; - C / a (x + B / 2a) &# 178; = (B & # 178; - 4ac) / 4A & # 178; when B & # 178; - 4ac0, there are two unequal real roots X1 = - B / 2A + √ (B & # 178; - 4ac) / 2A

Solving quadratic equation AX ^ 2 + BX + C = 0 (a is not equal to 0) with collocation method

(x+b/2a)^2-b/2+c

Let x1. X2. Be two of the quadratic equation of one variable 2x square + 4x-3 = 0. Find: x2 divided by the square of X1 plus X1 divided by the square of x2

According to Weida's theorem:
X1+X2=-2,X1*X2=-3/2,
X1^2+X2^2=(X1+X2)^2-2X1*X2=4+3=7
(X2/X1)^2+(X1/X2)^2
=［(X1^2+X2^2)^2-2(X1*X2)^2］/(X1*X2)^2
=7÷(9/4)-2
=28/9-2
=10/9

The relationship between X1: x2 and Weida's theorem
Such as the title
The value of X1: X2 is expressed by Weida theorem
It seems to have something to do with X1: x2 + x2: x1

x1:x2+x2:x1=(x1+x2)/x1x2
On the right, we can use the Veda theorem
On the left is the sum of the formula and its reciprocal
Then solve a quadratic equation

How to use Weida theorem to solve X1 / X2 (x1 / x2)

X2 / X1 + X1 / X2 is solved by Weida theorem

The original formula = (x1 & sup2; + x2 & sup2;) / x1x2
=[(x1+x2)²-2x1x2]/(x1x2)²
Let X1 + x2 = - B / A
x1x2=c/a
Just insert it

If the value of cube + 3bx + 4 of algebraic formula 2aX is 5 when x = 1, then the value of cube + 3bx + 4 of algebraic formula 2aX is 5 when x = - 1
If M-N = 5, Mn = - 2, then the square of (n-m) - 4Mn=

When x = 1,
The cube of 2aX + 3bx + 4 = 2A + 3B + 4 = 5
Then, 2A + 3B = 1
When x = - 1,
Cube of 2aX + 3bx + 4 = - 2a-3b + 4 = - (2a + 3b) + 4 = - 1 + 4 = 3
m-n=5,mn=-2,
Square of (n-m) - 4Mn
=The square of (- 5) - 4 × (- 2)
=25+8
=33

If x = 1, the value of cube + 3bx + 4 of algebraic formula 2aX is 5, then when x = - 1, the value of cube + 3bx + 4 of algebraic formula 2AR is 5
Write clearly the thinking and process

Substituting x = 1 into 2A + 3B + 4 = 5, that is, 2A + 3B = 1
X = - 1
-2a-3b+4
=-(2a+3b)+4
=-1+4
=3
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The square of a - 3AB + 9b, where a = - 2, B = two-thirds, and 2 (X-Y) - 3 (Y-X), where X-Y = - 1

Solution
a²-3ab+9b²
=(-2)²-3×(-2)×(2/3)+9×(2/3)²
=4+4+4
=12
2(x-y)-3(y-x)
=2(x-y)+3(x-y)
=5(x-y)
=5×(-1)
=-5