Decompose 3x ^ 2 + 7x + 2 Ask,

Decompose 3x ^ 2 + 7x + 2 Ask,

A:
1) Cross phase multiplication: 3 * 2 + 1 = 7
3x^2+7x+2=(3x+1)(x+2)
2）
3x^2+7x+2
=3x^2+6x+x+2
=3x(x+2)+(x+2)
=(3x+1)(x+2)

It is known that the equation AX & sup2; - 1 / 3 of X and b-2-2 / 3 of x = 0 are linear equations of one variable. Try to find the value of (a + b) of X

Because the equation AX & # 178; - 1 / 3 x b-2-2 / 3 = 0 about X is a linear equation of one variable
∴a=0,b-2=0
∴b=2
So the equation is 1 / 3x-2 / 3 = 0
So x = 2
So the (a + b) power of x = 2 & # 178; = 4

It is known that (2a-4) x & sup2; - ax + 8 = 0 is a linear equation of one variable with respect to X. (1) try to determine the value of a; (2) find the solution of this equation

2*a-4=0,a=2;
-2*x+8=0,x=4.

Sufficient conditions for at least one real root of equation AX2 + 2x + 1 = 0

The equation AX2 + 2x + 1 = 0 has at least one real root, Δ > = 0, Δ = 4-4a = 4 (1-A) > = 0 a

If the solution of inequality 2x < 4 can make the linear inequality (A-1) * x < A + 5 with respect to x hold, then the value range of a is I read your answer
We know that the root of equation 5 (x-a) = - 2A about X is greater than that of equation 3 (x-a) = 2 (x + a) about X, then a should be a. non zero number B is greater than - 1 number C negative number d positive number the correct answer is B, but how come?

As for the first question, you should tell me which step I can't understand, and then discuss it
In the latter problem, by solving equation 1, x = 3A / 5
By solving equation 2, x = 5A
Because the root of the former is greater than that of the latter
So 3A / 5 ＞ 5a, so a ＜ 0, that is to say, choose C

If the equation MX2 + (2m + 1) x + M = 0 of X has two unequal real roots, then the value range of M is ()
A. M ＞ - 14b. M ＜ - 14C. M ＞ 4D. M ＞ - 14 and m ≠ 0

The equation MX2 + (2m + 1) x + M = 0 about X has two unequal real roots. The equation is a quadratic equation with one variable, and △ = (2m + 1) 2-4m · m ＞ 0 and m ≠ 0, 4m2 + 1 + 4m-4m2 ＞ 0, 4m ＞ - 1, M ＞ - 14 and m ≠ 0

The two real number roots of the equation MX2 − (m − 4) x + M4 = 0 of X are X1 and x2. (1) if the equation has two unequal real number roots, find the value range of M; (2) is there a value of m such that X1 and X2 satisfy 1x1 + 1x2 = 0? If it exists, find the value of M. if it does not exist, explain the reason

(1) For the discriminant of ∵ root, the equation has two unequal real roots: ∵ △ 0, ∵ [- (M-4)] 2-4m · M4 > 0, ∵ (M-4) 2 > m2, ∵ m2-8m + 16 > m2; ∵ m < 2. (2) ∵ X1 + x2 = m − 4m; x1x2 = M4M = 14. ∵ 1x1 + 1x2 = X1 + x2x1x2 = m − 4m14 = 4 (m − 4) m; and ∵ 1; 1x1 + 1x

The equation x2 + (M-3) x + M = 0 of X has two unequal real roots in (0, 2), then the value range of M is ()
A. 23 < m ≤ 1b. 23 < m < 1C. 1 < m < 3D. M < 1 or M > 9

Let f (x) = x2 + (M-3) x + m, then △ = (m − 3) 2 − 4m ＞ 0f (0) = m ＞ 0f (2) = 4 + 2 (m − 3) + m ＞ 00 ＜− m − 32 ＜ 2, the solution is 23 ＜ M ＜ 1, so B is chosen

If the square of equation x-5x = a has only two different real roots, then the value range of a is PS: the answer is that a is greater than 25 ＼ 4 or a = 0

Deformation: a = | (X-5 / 2) ^ 2-25 / 4|
Take the right side of the equation as a function and draw a graph (it's a parabola, but the part less than 0 is on the x-axis)
The left side of the equation is equivalent to y = a
As long as there are two intersections on both sides of the equation, there are two different real roots
When a = 25, there are three points of intersection (that is, the maximum value), when 0

If a and B are the two real roots of the equation x plus 99x-1 = 0, then AB square plus a square B minus ab

x²+99x-1=0
∵ A and B are the square of the equation x plus the two real roots of 99x-1 = 0
∴a+b=-99
ab=-1
∴ab²+a²b-ab=ab(b+a-1)=(-1)×(-99-1)=(-1)×(-100)=100