Elementary two problems of quadratic equation with one variable It is known that ax ② + BX + C = 0 has two roots, and a + C = 0 can find the relationship between the two roots

Elementary two problems of quadratic equation with one variable It is known that ax ② + BX + C = 0 has two roots, and a + C = 0 can find the relationship between the two roots

Ax ^ 2 + BX + C = 0, a + C = 0, C = - A
ax^2+bx-a=0
According to Weida's theorem:
x1+x2=-b/a;
x1*x2=-1;
According to the product of two equal to - 1, that two reciprocal negative relationship
According to Weida's theorem: X1 * x2 = C / a = - 1
So X1 = - 1 / x2
According to Weida's theorem, X1 * x2 = C / A
Because a + C = 0
So a = - C
So X1 * x2 = - 1
So X1 = - 1 / x2
Ax^2+Bx+C=0,
A + C = 0, that is, when x = 1, a + B + C = 0, B = 0,
X1+X2=-B/A=0,
X1 * x2 = C / A, and a + C = 0, a = - C,
X1*X2=-1.
The relationship between the two is: X1 * x2 = - 1, X1 + x2 = 0
8y-1 = 4Y square by CM method
8y-1=4y²
4y²-8y=-1
4y²-8y+4=3
4(y-1)²=3
(y-1)²=3/4
y-1=±√3/2
y=1±√3/2
7x²=5
7x²-2x+5=0
So it is
The square of Y + 8y + 1 = 0
The square of Y + 8y + 1 = 0
The square of Y + 8y + 16 = 15
(y+4)²=(√15)²
Y = - 4 + √ 15 or y = - 4 - √ 15
Help solve an equation: 8y + (the square of 2Y)
Is 8y + (the square of 2Y) = 4
8y+4y²=4
y²+2y-1=0
Y = - 1 ± root 2
Let 2 square of equation x plus 2 XY minus 2 square of 8y equal to 0 be transformed into two linear equations________ .
x^2+2xy-8y^2 = (x+4y)(x-2y) =0
That is x + 4Y = 0
Or x-2y = 0
$(word)
x^2+2xy-8y^2=o
(x+4y)(x-2y)=0
Cross multiplication
1. Linear equation
(1) point oblique:; (2) oblique section:; (3) intercept:;
(4) two point formula: (5) general formula: (a, B are not all zero)
(1) if the slope of a straight line is known to be K, and if it passes through a point (x0, Y0), the equation of a straight line is y-y0 = K (x-x0). (2) if the slope of a straight line is known to be K, and the ordinate of its intersection with the y-axis is B, the equation of a straight line is y = KX + B. (3) if the abscissa of its intersection with the x-axis is a, and the ordinate of its intersection with the y-axis is B, the equation of a straight line is
How about the coefficient of division by the base power, such as the second power of 2x △ the second power of X
The coefficients are also divided
2X to the second power of x = 2
Given that ab satisfies a & # 178; + B & # 178; + 4a-8b + 20 = 0, try to decompose the factor (X & # 178; + Y & # 178;) - (B + ax)
And we're going to be 178, + B & \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\35;178;) - (B + axis) = (X & #
The former formula can be reduced to: (a + 2) ^ 2 + (B-4) ^ 2 = 0, obviously, a = - 2, B = 4
So substituting into the second formula, we get: (x + y) ^ 2-4 = (x + y + 2) (x + Y-2)
A factoring problem in high school mathematics
How to factorize x ^ 2 + (m-1) x-m?
Cross multiplication
1 m
1 -1
So, the original formula = (x + m) (x-1)
（x+m)(x-1)
Factorization is (x + m) (x-1)
Cross phase multiplication, get (x + m) (x-1)
(x-1) (x + m) is OK
If y = x & # 178; - 3x + 2 and Y axis intersection coordinates are
If y = - 2x & # 178; + 5x-3 axis intersection coordinate is
A:
The point of intersection of Y-axis and parabola
When x = 0, you get y = 2
The intersection coordinates of the parabola y = x & # 178; - 3x + 2 and the Y axis are (0,2)
The intersection coordinates of the parabola y = - 2x & # 178; + 5x-3 and the Y axis are (0, - 3)