(x + a) (x + b) = the square of x-13x + 36, find the value of a + B

(x + a) (x + b) = the square of x-13x + 36, find the value of a + B

：（x+a)(x+b)=x²＋﹙a＋b﹚x＋ab＝x²－13x＋36
∴a＋b＝－13

It is known that the analytic formula of quadratic function is y = ax square-3 + 2x
1) When you ask what value a takes, the function image has an intersection with the X axis
2) If the x-axis of the image graph of this quadratic function intersects at two points a (x1,0) and B (x2,0), when AB = 4, find a
The answer is (√ 4 + 12a) / a = 4

(1) If a function has an intersection with the x-axis, it means that the quadratic equation of one variable has a real root, that is, the discriminant ≥ 0,
So, 2 ^ 2 + 12a ≥ 0, a ≥ - 1 / 3, in addition, the quadratic function a ≠ 0, so the value of a is a ≥ - 1 / 3 and a ≠ 0
(2) AB = 4, we can use the formula of distance between the intersection of quadratic function and horizontal axis: | ab | = [√ (b ^ 2 - 4ac)] / | a | = 4
That's the step you asked. The rest is to solve the equation. But a on the denominator should be signed with absolute value
（√4+12a）/|a|＝4
（√4+12a）=4|a|
4+12a=16a^2
4a^2-3a-1=0
(4a+1)(a-1)=0
A = - 1 / 4 or a = 1

Is (x's Square minus 2x minus 2) multiplied by (x's Square minus 2x plus 4) plus 9 a complete square formula? Why? Write the formula

The original formula = [(X & sup2; - 2x) - 2] [(X & sup2; - 2x) + 4] + 9
=(x²-2x)²+2(x²-2x)-8+9
=(x²-2x)²+2(x²-2x)+1
=(x²-2x+1)²
So it's a perfect square

How can the equation (8-2x) (5-2x) = 18 be solved to 2x square - 13X + 11 = 0

Multiply it
40-16x-10x+4x²=18
4x²-26x+22=0
Divide by two
2x²-13x+11=0
(2x-11)(x-1)=0
x=11/2,x=1

The solutions of binary linear equations ({5x + y = 84, {6x + 3Y = 108) are obtained

5x+y=84 (1)
6x+3y=108 (2)
(1)×3-(2)
15x+3y-6x-3y=252-108
9x=144
x=16
y=84-5x=4

Given the equation 5x + 3Y = 22, find the nonnegative integer solution of the equation,

5x+3y=22
y=(22-5x)/3=7-2x+(1+x)/3
Then (1 + x) / 3 is an integer
Let a = (1 + x) / 3
x=3a-1
y=(22-15a+5)/3=9-5a
x>=0,y>=0
So 3a-1 > = 0, a > = 1 / 3
9-5a>=0,a

Solving the quadratic equation of 2 variables 5x-3y = 19 6x-5y = 27
Writing process by substitution

X=（3y+19）÷ 5
Substituting formula 2
6[（3y+19）÷ 5] - 5y =27
Is it good to calculate y = by yourself?

The solution of the equation 5x-1-a = 0 about X is 2 smaller than that of the equation 3Y + 1 + a = 0 about y

From 5x-1-a = 0
Then a = 5x-1
From 3Y + 1 + a = 0
Then a = - (3Y + 1) = - 3y-1
Then 5x-1 = - 3y-1,
And from x + 2 = y,
So 5x-1 = - 3 (x + 2) - 1,
That is, 8x = - 6,
X=-3/4.
So a = 5x (- 3 / 4) - 1 = - 19 / 4

Equation solution: 240% x-x = 6 1 / 4 + 25% x = 2 / 3 2.5 * 40% x-3 / 5x = 4
Quick, quick
Subject to the following, there is a mistake: 240% x-x = 6 1 / 4 + 25% x = 3 / 2 2.5 * 40% x-3 / 5x = 4

240％X-X=6
1.4x=6
x=30/7
1/4+25％X=2/3
1/4x=2/3-1/4
1/4x=5/12
x=5/3
2.5*40％X-3/5X=4
2/5x=4
x=10

Equation x + 5x of 9 = x of 18
Equation x + 5 / 9 x = 1 / 18

Original formula: 5x / (x + 9) = x / 18
x^2+9x=90x
x^2-81x=0
x(x-81)=0
After inspection
The solution is x = 0 or x = 81