Given that the square of X + ax + 4 is a complete square, then the value of a is I know it's 4 or - 4 Please help me explain why

Given that the square of X + ax + 4 is a complete square, then the value of a is I know it's 4 or - 4 Please help me explain why

Because the number term (4) of the complete square formula is equal to the square of a number, the number is + - 2. From the form of the complete square formula x2 + 2XY + Y2, where y is positive and negative 2, so a is equal to 2Y, which is positive and negative 4

If the algebraic formula x square-ax + 2a-1 is a complete square, what is the value of a?

If X & # 178; - ax + 2a-1 is a complete square, then
The complete square is (x-a / 2) &# (178);
So a & # / 4 = 2a-1
a²=8a-4
a²-8a+4²=-4+16
(a-4)²=12
a-4=±√12
a=4±2√3

If the square of X + ax + B is a complete square, the relationship between a and B is

The square + ax + B of X is a complete square formula, which can be written as follows:
The square of X + ax + B = (x + k) &# 178; = x & # 178; + 2K + K & # 178;,
2k=a,k²=b,
That is, k = A / 2, K & # 178; = (A / 2) &# 178; = B,
The relationship between a and B is b = (A / 2) &

If the quadratic trinomial 4x & # 178; ax + 1 is a complete square, then the value of a is a.4 B. plus or minus 4 C. - 4
A.4
B. Plus or minus 4
C.-4
We need reasons.

When the 4x & #178; + ax + 1 is a complete square 4x & \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\thesolution is: a = 4

If the quadratic trinomial x2 + AX-1 can be decomposed into (X-2) (x + b), then the value of a + B is ()
A. -1B. 1C. -2D. 2

(X-2) (x + b) = x2 + (b-2) x-2b, ∵ quadratic trinomial x2 + AX-1 can be decomposed into (X-2) (x + b), ∵ a = B − 2 − 2B = − 1, the solution is a = − 32B = 12, ∵ a + B = - 32 + 12 = - 1

When x = -- 4, the square of the quadratic trinomial ax -- 4x -- 1 is - 1. When x = 5, the value of the quadratic trinomial is obtained

Ax ^ 2-4x-1 into x = - 4
=a*(-4)^2-4*(-4)-1
=16a+16-1=16a+15=-1 => a=-1
That is, the original formula is - x ^ 2-4x-1 when x = 5
Original formula = - 5 ^ 2-4 * 5-1
=-25-20-1=-46

The cube of the square X of a + the square of 2x - the cube of 4x + the square of 2x - ax - 9 is a quadratic trinomial

a²x³+2x²-4x³+2x-ax²-9
=(a²-4)x³+(2-a)x²+2x-9
∵ is a quadratic trinomial
Ψ A & # 178; - 4 = 0 and 2-A ≠ 0
∴a=-2
The quadratic trinomial is: (2-A) x & # 178; + 2x-9

Complex number solution equation
Solve the equation AX ^ 2 + BX + C = 0 about X in the complex number range, where a, B and C are real numbers

If B ^ 2-4ac = 0, then x = - B / 2A
If B ^ 2-4ac > 0, then x = [(b ^ 2-4ac)] / (2a)
If B ^ 2-4ac

Solving the equation SHZ = I
The answer is ZK = x + iy = (2k π + 0.5 π) I

Because SHZ = (e ^ Z-E ^ - 2) / 2, this is a formula, then e ^ Z-E ^ - z = 2I, and then using the formula e ^ z = e ^ x (cosy + isiny) to simplify, e ^ 2x (cos2y + isin2y) - 2ie ^ x (cosy = isiny) - 1 = 0, x = 0, then cos2y + sin2y + isin2y-2icosy-1 = 0, that is, sin2y-2cosy = 0 and cos2y + 2Si

complex
0 - resolution time: 2008-9-20 17:16
z^2-3(1+i)z+5i=0
To solve this equation,

Let z = a + bi, a and B be real numbers, then Z ^ 2-3 (1 + I) Z + 5I = 0 (a + bi) ^ 2-3 (1 + I) (a + bi) + 5I = 0A ^ 2-B ^ 2-3a + 3B + (2ab-3b-3a + 5) I = 0, that is, a ^ 2-B ^ 2-3a + 3B = 02ab-3b-3a + 5 = 0. The solution of the equations shows that a = 1, B = 2 or a = 2, B = 1, so z = 1 + 2I or Z = 2 + I