The relation between the root and coefficient of quadratic equation of one variable, X & # 178; + X + 4N = 0 and 2x & # 178; + 7x + 3N = 0 have a common root, where n is not equal to 0. Try to find the value of their roots (not expressed in n)

Let the common root be x1
X & # 178; + X + 4N = 0 are X1 and x2
2X & # 178; + 7x + 3N = 0 are X1 and X3
According to the relationship between root and coefficient
x1+x2=-1
x1*x2=4n
x1+x3=-7/2
x1*x3=3n/2
The first formula subtracts the third formula to get x2-x3 = 5 / 2
Because n is not equal to 0, we can divide formula 2 by formula 4 to get x2 / X3 = 8 / 3
So X3 = 3 / 2, X2 = 4
Finally, we get X1 = - 5

The relationship between the root and coefficient of quadratic equation of one variable
(1) Given the quadratic equation of one variable x square - (M + 1) x - (m Square-1) = 0, if the two real roots of the equation are x1, X2 and satisfy the condition that one part of X1 + one part of x2 = minus two-thirds, find the value of M and two parts of the equation (2) given that the two roots of the equation X2 - (2a-1) x + 4 (A-1) = 0 of X are the lengths of the two right sides of the right triangle with hypotenuse 5, find the area of the right triangle

(1) Because 1 / X1 + 1 / x2 = (x1 + x2) / x1x2
According to the relationship between the root and coefficient of quadratic equation
x1+x2=m+1,x1x2=-m^2+1
So 1 / X1 + 1 / x2 = (M + 1) / (- m ^ 2 + 1) = (M + 1) / - (M + 1) (m-1) = 1 / (1-m) = - 2 / 3
-2+2m=3
m=5/2
The equation is x ^ 2-7x / 2-21 / 4 = 0
Then solve the equation
(2) According to the relationship between the root and coefficient of quadratic equation
x1+x2=2a-1,x1x2=4(a-1)
From Pythagorean theorem
x1^2+x2^2=5^2
(x1+x2)^2-2x1x2=25
So (2a-1) ^ 2-8 (A-1) = 25
4a^2-12a+9=25
4a^2-12a-16=0
a^2-3a-4=0
A = 4 or a = - 1
When a = 4, the equation is x ^ 2-7x + 12 = 0
X = 3 or x = 4, area = 3 * 4 / 2 = 6
When a = - 1, the equation is x ^ 2 + 3x-8 = 0, and the equation has negative roots
So the area is 6

As shown in the figure, there is a small bulb with a rated voltage of 2.4V. The resistance of the small bulb is 8 Ω when it works normally. If we only have a 3V power supply, how large a resistance needs to be connected in series to make the small bulb work normally?

The current in the circuit I = ulrl = 2.4v8 Ω = 0.3A, the voltage at both ends of the resistor ur = u-ul = 3v-2.4v = 0.6V, so the resistance value of the series resistor R = URI = 0.6v0.3a = 2 Ω. A: a 2 Ω resistor needs to be connected in series

Find the value of 49x & # 178; - 16 = 0, 8x & # 179; + 125 = 0

49x²-16=0
X^2=16/49
X=±4/7
8x³+125=0
X^3=-125/8
X=-5/2.

The electromotive force of the power supply is 1.5V, the resistance is 0.12 ohm, and the external resistance is 1.38 ohm

Current = 1 A, 1.5 / (0.12 + 1.38)
Terminal voltage [voltage on external resistance] 1.38V

Hyperbola x ^ 2-y ^ 2 / 4 = 1, the line L passing through P (1,1) has only one common point with hyperbola, so the equation of L is obtained
My answer is so strange

Let the equation of l be Y-1 = K (x-1), and then it is combined with hyperbolic equation to obtain a quadratic inequality of one variable. Because there is only one common point, the discriminant of the root is 0, and the equation of K is solved. Then consider the case that K does not exist, and draw a picture, because the equation of L is x = 1
The method tells you to do it yourself

For copper and iron wires of the same thickness, when the temperature is the same, their resistance is the same, then ()
A. Copper wire is longer than iron wire B. iron wire is longer than copper wire C. the length of two copper iron wires is the same D. It is impossible to judge

Copper and iron wires with the same thickness have the same resistance when the temperature is the same. Because copper has better conductivity than iron, the length of copper wire should be longer than iron wire. Therefore, options B, C and D are incorrect. Therefore, option a is selected

It is known that ellipse D: x250 + y225 = 1 and circle M: x2 + (Y-5) 2 = 9, hyperbola g and ellipse d have the same focus, and its two asymptotes are just tangent to circle m. The equation of hyperbola G is obtained

Let the equation of hyperbola G be x2a2-y2b2 = 1 (a ﹥ 0, B ﹥ 0) ﹥ asymptote be BX ± ay = 0 and A2 + B2 = 25; the distance between the center of circle m (0,5) and the two asymptotes is r = 3, ﹤ 5A | A2 + B2 = 3, that is, 5 ﹤ a | 5 = 3; the solution of a = 3, B = 4, ﹤ g equation is x29-y2 16=1．

The power written on the heater is 1500W. How many kilowatts of electricity can you use in an hour?
The rated power is 1500W

If you drive to the top gear, the power consumption is 1.5 degrees an hour. If it's not in the top gear, it's less than 1.5 degrees

1 is that the tangent equation of the curve at the point (x, f (x)) of FX is y = KX + B, where the slope k stands for? Is it equal to the derivative of FX? B? 2 in the mathematical induction method, when I set n = k, the result proves that n = K + 2 also holds, but n = K + 1 does not hold,

1. The slope k represents the derivative of F (x) when x = x0, that is, k = f '(x0)
The tangent equation of F (x) at point (x0, f (x0)) is y-f (x0) = f '(x0) (x-x0)
It is reduced to y = f '(x) · x + F (x0) - f' (x0) · x0
∴b=f(x0)-f'(x0)·x0
2. No. mathematical induction must prove that n = K + 1 is tenable. Others are not

The relation between the root and coefficient of quadratic equation of one variable, X & # 178; + X + 4N = 0 and 2x & # 178; + 7x + 3N = 0 have a common root, where n is not equal to 0. Try to find the value of their roots (not expressed in n)