When k takes any value, the roots of the quadratic equation (k2-1) x2 + (3k-9) x = 18 are integers When k takes any value, the roots of the quadratic equation (K & # 178; - 1) x & # 178; + (3k-9) x = 18 are integers

When k takes any value, the roots of the quadratic equation (k2-1) x2 + (3k-9) x = 18 are integers When k takes any value, the roots of the quadratic equation (K & # 178; - 1) x & # 178; + (3k-9) x = 18 are integers

When k ≠ - 1 and K ≠ 1, the solution is x = - 6 / (K + 1) or x = 3 / (k-1) because x is an integer, so K + 1 is a divisor of 6 and k-1 is a divisor of 3

Solving quadratic equation x ^ 2-6x + 5 = 0

x^2-6x+5=0
(x-5)(x-1)=0
x1=5
x2=1

(x + 1) (x + 8) = - 12 univariate quadratic equation

(X＋1)(X＋8)＝－12
x²+9x+8=-12
x²+9x+20=0
（x+4)(x+5)=0
X = - 4 or x = - 5

To solve quadratic equation of one variable: (1) 3x2-6x + 1 = 0 (collocation method) (2) x2 + 3x = 2 (formula method) (3) 3x (x + 2) = 5 (x + 2) (4) (x + 8) (x + 1) = - 12

(1) 3x2-6x + 1 = 0 (matching method) x2-2x = - 13, x2-2x + 1 = 1-13, (x-1) 2 = 23, X-1 = ± 63, X1 = 1 + 63, X2 = 1-63; (2) x2 + 3x = 2 (formula method), X2 + 3x-2 = 0, a = 1, B = 3, C = - 2, b2-4ac = 9 + 8 = 17 & gt; 0, x = - B ± b2-4ac2a = - 3 ± 172 × 1,  X1 = - 3 +

20 + 20 (x + 1) + 20 (x + 1) ^ 2 = 72.8 Chus elementary ternary quadratic equation

That is, 1 + (x + 1) + (x + 1) = 3.64 x + 3x-0.64 = 0 100x + 300x-64 = 0 25X + 75x-16 = 0 (5x-16) (5x + 1) = 0, so 5x-16 = 0 or 5x + 1 = 0, so x = 16 / 5 or - 1 / 5

Solving quadratic equation of one variable: (x ^ 2 + x) (x ^ 2 + X + 2) - 8 = 0
Advice to individual responders: copy shamelessly

Let x ^ 2 + x = y
The original formula = y (y + 2) - 8 = y ^ 2 + 2y-8 = (y + 4) (Y-2) = 0
Y = - 4 or y = 2
So x ^ 2 + X + 4 = 0 or x ^ 2 + X-2 = 0
When x ^ 2 + X + 4 = 0, the equation △ = 1-16 = - 15

If only one of the two quadratic equations x2 + NX + M = 0 of X is equal to 0, then the following condition is correct ()
A. m=0，n=0B. m≠0，n≠0C. m≠0，n=0D. m=0，n≠0

∵ only one of the two quadratic equations x2 + NX + M = 0 is equal to 0, ∵ X1 + x2 = - n ≠ 0, x1x2 = M = 0, ∵ M = 0, n ≠ 0

A mathematical problem, one variable quadratic equation: x ^ 2-14x = 8, find X
X ^ 2-14x = 8, find the value of X

How to use and match
X^2-14X=8
（x-7）²-49=8
（x-7）²=57
So x = (7 ± root 57)

What is the relationship between the root of the quadratic equation x of one variable and the image of the quadratic function y = x of square-6x + 3

The root of the quadratic equation x with one variable is the abscissa of two points whose ordinate is 11 on the image of the quadratic function y = x with square - 6x + 3
The method of drawing on the image is: take a point on the y-axis whose coordinates are (0,11), make a parallel line of x-axis through this point, and there are two intersections with the parabola. The number corresponding to the intersection of the vertical line of x-axis and X-axis through these two intersections is the root of the equation

5x (X-2) - x + 2 = 0, the simplest way is to solve the quadratic equation with one variable,

5x(x-2)-x+2=0
The deformation results are as follows
5x(x-2)-(x-2)=0
Extracting the common factor X-2, we get
(x-2)(5x-1)=0
Can get
X = 2 or x = 1 / 5