The sixth power of (- y) - (- 3Y & # 179;), # 178; - [- (2Y) &# 178;], # 179;

The sixth power of (- y) - (- 3Y & # 179;), # 178; - [- (2Y) &# 178;], # 179;

The sixth power of (- y) - (- 3Y & # 179;), # 178; - [- (2Y) &# 178;], # 179;
=The sixth power of Y - the sixth power of 9y + the sixth power of 8y
=0

Use the formula method to solve the following equation: x ^ 2 + 2x-2 = 0
thx

This is a quadratic equation of one variable
Quadratic coefficient a = 1
Coefficient of first order term B = 2
The constant term is C = - 2
The root formula is: x = (- B ± B ^ 2-4ac) / 2A
Substitute the above values of a, B and C into the formula
We can get: x = (- 2 ± 2 ^ 2-4 * 1 * (- 2)) / 2
X = (- 2 ± 12 under root) / 2
X = (- 2 ± 2 times root 3) / 2
X = - 1 ± root 3
So:
X1 = - 1 + radical 3
X2 = - 1-radical 3

The formula method of solving the equation 2x ^ 2-mx-n ^ 2 = 0 about X
Thank you

a=2,b=-m,c=-n^2
So the discriminant △ = B ^ 2-4ac = m ^ 2 + 8N ^ 2
So X1 = [M - √ (m ^ 2 + 8N ^ 2)] / 4, X2 = [M + √ (m ^ 2 + 8N ^ 2)] / 4

Using the formula: (3-y) ^ 2 = 2Y (Y-3), 3x ^ 2 - (x + 2) ^ 2 + 2x = 0

1.
(3-y) ^ 2 = (Y-3) ^ 2, so y = 0 or 2Y = Y-3, y = - 3
two
3x ^ 2 - (x + 2) ^ 2 + 2x = 3x ^ 2-x ^ 2-4x-4 + 2x = 2x ^ 2-2x-4 = 2 (x + 1) (X-2) = 0, so
X = - 1 or 2

Taking 2 + √ 3 and 2 - √ 3 as roots, the univariate quadratic equation with quadratic coefficient of 1 is?

x^2-4x+1=0

x+2/5x=9/20

7/5x=9/20
x=9/20÷7/5
x=9/28

(1) Given the equation x ^ 2 + X-1 = 0, find a quadratic equation with one variable so that its roots are the opposite of the known roots
The equation is: ()
(2) It is known that the quadratic equation AX ^ 2 + BX + C = 0 (a ≠ 0) with one variable has two real roots which are not equal to zero. Find a quadratic equation with one variable so that its roots are the reciprocal of the known equation

(1) The roots of the new equation are the opposite of the known roots,
Then the product of two new equations is equal to that of two known equations;
The sum of two new equations is the opposite of the sum of two known equations
The equation is: x ^ 2-x-1 = 0
(2) Let two of the original equations be: x1, x2
Then the two equations are: 1 / x1, 1 / x2
x1x2=c/a x1+x2=-b/a
1/x1*1/x2=1/(c/a)=a/c
1/x1+1/x2=(x1+x2)/(x1x2)=(-b/a)/(c/a)=-b/c
The equation is: CX ^ 2 + BX + a = 0

How to do 5x + 2 (8-x) = 20

Original formula: 5x + 16-2x = 20
3X=4
X=3/4

Given the equations x + y = a; xy = B, X and y can be regarded as quadratic equations of one variable with respect to Z________________ Two solutions of the problem

Suppose the equation is
z^2+mz+n=0
By Weida theorem
x+y=-m,xy=n
So a = - m, B = n
m=-a,
So Z ^ 2-az + B = 0

2.5x x 6 (X-19) = 90

-24/17