If two univariate quadratic equations XX + X + a = 0 and XX + ax + 1 = 0 with real coefficients about X have a common real root, then a= If two quadratic equations XX + X + a = 0 and XX + ax + 1 = 0 with real coefficients of X have a common real root, what is a?

If two univariate quadratic equations XX + X + a = 0 and XX + ax + 1 = 0 with real coefficients about X have a common real root, then a= If two quadratic equations XX + X + a = 0 and XX + ax + 1 = 0 with real coefficients of X have a common real root, what is a?

∵ there is a common root, so this root is the root of these two equations, that is, the root of the following equations
x²+x+a=0 (1)
x²+ax+1=0 (2)
(1) - (2) get
(1-a)x=1-a
When 1-a0, there is a root x = 1
When a = 1, there are innumerable roots x (the two equations are the same)
When x = 1, a = - 2

As shown in the figure, in a square ABCD with a side length of 3cm, circle O1 is circumscribed with circle O2, circle 01 is tangent to Da and DC, and circle 02 is tangent to Ba and BC. Calculate the center distance between two circles

In addition, O1 and AD are cut to e, O2 and B are cut to CF, and O1, e, O2 and F are connected
O1E ⊥ AD,

x. If y satisfies (10y-3x) / (3x-y) = 2 and X is greater than 0, (3x ^ 2-2y ^ 2) / (5Y ^ 2-x ^ 2)=

It is necessary to de denominator (10y-3x) / (3x-y) = 2
(4/3)y=x
What is the formula
(3x^2-2y^2)/(5y^2-x^2)
=30/29

Solving the equation 3x + 2Y = 10
Natural number solution

From 3x + 2Y = 10, we get
Y = 5 + (- 3x / 2) x can only take even numbers from natural numbers
When x = 0, y = 5
When x = 2, y = 2
When x = 4, y = - 1 is not suitable
Therefore, the solution of equation 3x + 2Y = 10 has two groups of x = 0, y = 5 and x = 2, y = 2

How to solve x + y = 10 and 3x + 2Y = 50-26

The answer is x = 4, y = 6

Given the power of 10x = 2, the power of 10Y = 3, the power of 10z = 5, find the value of 3x + 2y-z of 10

3x + 2y-z of 10
=(10^x)³(10^y)²/10^z
=8×9/5
=72/5

(1) 6y = y (2) 5 = 3x, make them into general form

（1）6y-y =0（2）3x -5=0

When x + y = 2, - 3x-6y + 5 =?

-3x - 6y + 5
= -3(x + 2y) + 5
= -3×2 + 5
= -6 + 5
= -1

If X - (5 + 2Y) = 12, then 3x-6y=

Because X - (5 + 2Y) = 12, so x-5-2y = 12, so x-2y = 17, so 3x-6y = 17 × 3 = 51

{3x-4y = 10 ① 5x + 6y = 12 ② solve the equation

①*5 15x-20y=50
②*3 15x+18y=36
And then subtract the two
38y = - 14
y=-7/19
x=54/19