The proof of X & sup2; + (K + 2x) + 2K = 0, no matter K takes any real number, the root of the square is also a real number

The proof of X & sup2; + (K + 2x) + 2K = 0, no matter K takes any real number, the root of the square is also a real number

Is the title x2 + (K + 2) x + 2K = 0?
Discriminant = (K + 2) ^ 2-8k = (K-2) ^ 2 > = 0
So there's always a real solution
Hope to adopt, thank you

If the equation x ^ 2-4|x| + 5 = m of X has at least three real roots, then the value range of real number m is

The original equation is transformed into standard form
x²－4|x|＋(5－m)＝0
Namely
When x ≥ 0, it is X & # 178; - 4x + (5-m) = 0
When x < 0, it is X & # 178; + 4x + (5-m) = 0
According to Weida's theorem, △ 0
That is 4 & # 178; - 4.1 · (5-m) > 0
∴m < -1

The solution of the equation SiNx = (2 / Π) x

First of all, Y1 = SiNx y2 = (2 / Π) x, finding the abscissa of the intersection of Y1 and Y2 is the answer. Draw a picture to know that x = 0 is a solution, and then analyze whether there is focus. I don't know if you have learned derivative. The derivative of SiNx is cosx, which is the slope, and then Y2 is a positive proportional function

If the equation (M + 3) x ^ 2 + (2m + 5) x + M = 0 has two equal real roots, then M=____ And the root of the equation is____ .

If the equation (M + 3) x ^ 2 + (2m + 5) x + M = 0 has two equal real roots, then M=__ -25/8__ And the root of the equation is_ -10___ . △=(2m+5)^2-4(m+3)m =4m^2+20m+25-4m^2-12m =8m+25=0 m=-25/8 x=-b/2a=-(2m+5)/2(m+3)=-(-25/4+5)/2(-2...

If the equation m2x2 + (2m + 1) x + 1 = 0 of X has real roots, then the value range of M is ()
A. M ≥ − 14b. M ≥ − 14 and m ≠ 0C. M ≥ − 12D. M ≥ − 12 and m ≠ 0

When m = 0, the original equation can be reduced to x + 1 = 0, and the solution is x = - 1; when m ≠ 0, the equation m2x2 + (2m + 1) x + 1 = 0 about X has a real root, and the solution is m ≥ - 14, and the value range of M is m ≥ - 14

It is known that the equation x & sup2; + (2m-3) x + M & sup2; = 0 with respect to X has two unequal real number roots, and the value range of M is obtained

If the quadratic equation has two unequal real roots, then the discriminant △ 0
Namely
(2m-3)^2-4m^2＞0
(2m-3-2m)(2m-3+2m)＞0
-3（4m-3)＞0
But - 3 ＜ 0, 4 M-3 ＜ 0
→m＜3/4

How to prove that the equation x ^ 3-3x + 1 = 0 has and only has one root in the interval (0,1)?

y=x^3-3x+1
y'=3x^2-3
0

1. Prove that the equation E ^ x = 3x has at least one positive root less than 1

Let f (x) = e ^ x-3x
F '(x) = e ^ x-3, when 0 ＜ x ＜ 1, f' (x) ＜ 0, which means that f (x) is reduced
f(0)=1,f(1)=e-3

Given the function f (x) = x2 + 2cosx, then the sum of all real roots of the equation f (x) = f (x + 1x + 2) about X is ()
A. 0B. -2C. -4D. -6

The derivative function can be obtained as follows: F '(x) = 2x-2sinx, when x ≥ 0, f' (x) ≥ 0, ∵ f (- x) = f (x) increases on [0, + ∞), the ∵ function is even function, the ∵ equation f (x) = f (x + 1x + 2) is equivalent to x = x + 1x + 2 or x + X + 1x + 2 = 0 ∵ x2 + X-1 = 0 or x2 + 3x + 1 = 0 ∵ the sum of all real roots of the equation is - 4, so C is selected

It is proved that the equation 4x = 2 ^ X has and only has one real root on [0,1]

f(x)=2^x-4x
f'(x)=2^x*ln2-4
Obviously, f '(x) is increasing
And f (1) = 2ln2-4