It is known that the equation kx2-4kx + K-5 = 0 about X has two equal real roots. Find the value of K and solve the equation

It is known that the equation kx2-4kx + K-5 = 0 about X has two equal real roots. Find the value of K and solve the equation

∵ the original equation has two equal real roots, ∵ K ≠ 0 and △ = 0, that is, 16k2-4k (K-5) = 0, ∵ k = − 53 or K = 0 (rounding), ∵ the original equation can be reduced to − 53x2 + 203x − 203 = 0, ∵ 53 (x2 − 4X + 4) = 0, ∵ (X-2) 2 = 0, ∵ X1 = x2 = 2

Solve the equation x & # 178; + 9x-9 = 21

X²＋9X－9＝21
x^2+9x-30=0
From B ^ 2-4ac = 81 + 120 = 201
X = (- B ± root 201) / 2A = (- 9 ± root 201) / 2
So the solution is X1 = (- 9 + radical 201) / 2
X2 = (- 9-radical 201) / 2
I hope my answer can help you

Why X & # 178; + 4 / X & # 178; = 21?

x+2/x=5
Square on both sides
x²+2*x*2/x+4/x²=25
x²+4+4/x²=25
x²+4/x²=21

98765 out of 98766, 9876 out of 9877, 987 out of 988, and 88 out of 99 are arranged from small to large

88/99

Take e, F, G and h on the edges AB, BC, CD and Da of the space quadrilateral ABCD. If EF and Hg intersect at point m, then ()
A. M must be on line AC. B. m must be on line BD. C. m may be on line AC or on line BD. D. m is neither on line AC nor on line BD

Since ABCD is a quadrilateral in space, AB and BC determine the plane ABC, CD and Da determine the plane ACD. ∵ e ∈ AB, f ∈ BC, G ∈ CD, H ∈ Da ⊂ EF ⊂ plane ABC, GH ⊂ plane ACD ∩ GH = m ∩ m ∈ plane ABC, m ∈ plane ACD & nbsp; ∩ plane ABC ∩ plane ACD = AC ∩ m ∈ plane ACD, so select a

Among the five numbers, the operation symbols and brackets are added in the middle of the five numbers, and the operation symbols and brackets are added in the middle of the five numbers, so as to make the equation to be set as the formula of the five numbers, as you add the operation symbols and brackets, as well as the operation symbols and brackets in the middle of the five numbers, so as to add the operation symbols and brackets in the middle of the five numbers, so as to make the equation to be set as the formula of the formula is 0.0.0.1 & amp & nbsp; 1.1-1-1-1-1-1 & amp & nbsp; as as as as as as as as as as as you'0.0.1-1 & amp & amp & amp; as as as as as as as as as as as as as as as as as as as as as as as as as as as as as as as as as as as as as as as as as as as as as as as as as as as as as as as as as as as as as as as as as as as as as as as as as& nbsp; 0.4 & nbsp; & nbsp; 0.5 = 0.5

（0.1+0.2-0.3）×0.4×0.5=0 （0.1+0.2）÷0.3-0.4-0.5=0.1 （0.1+0.2+0.3）÷0.4-0.5=1 （0.1-0.2-0.3+0.4）×0.5=0

What is the relationship between the value of adjoint matrix of matrix A and the eigenvalue of matrix A?

Because a * a = iaie
IA*AI=IIAIEI=IAI^n,
IA*IIAI=IAI^n,
So ia * I = IAI ^ (n-1),
If a can be diagonalized, the eigenvalues of a are D1, D2,..., DN
Then IAI = D1D2,..., DN
So ia * I = IAI ^ (n-1) = (D1D2,..., DN) ^ (n-1)

A 4-digit number is equal to 24, and the number of single digits and hundred digits are even. Who can tell me what the number is?

The number of one digit is 0, the number of hundred digit may be (1) 6, the number is 9690 (2) 8, the number is 987088707890, the number of hundred digit may be (1) 4, the number is 9492 (1) 6, the number is 967286827692 (2) 8, the number is 985288627872,68

Let a and B be invertible matrices of order 5, a * be the adjoint matrix of a, and | a | = 0.5, | B | = 0.5, then the determinant | 3A * B | =?
Wrong | B | = - 0.5

|3A*B| = 3^5 |A*||B| = 3^5 |A|^4 |B| = 3^5 * 0.5^4 * (-0.5) = -243/32

Given the points a (- 1,2), B (3,4), find the point normal equation of the vertical bisector l of ab

Vector AB = (4,2)
The vertical bisector of AB must pass the midpoint of AB, and the midpoint of AB [(- 1 + 3) / 2, (2 + 4) / 2] = (1,3)
The point normal equation of the vertical bisector l of AB: 4 * (x-1) + 2 (Y-3) = 0