# Math from Beginner to Mastery: Quadratic Functions and Cross Multiplication

## Mathematics from Beginner to Master

### foreword

I wanted to learn mathematics well, but found that I may not remember the knowledge of junior high school, so I found a “Complete Works of High School Mathematics Fundamentals and Solutions” on station B again.

### Math Learning Path

  1 2 3  graph LR Math-.Function.->Quadratic function-.Solution.->Cross multiplication

### cross multiplication

This is the basis for the transition between junior high school and high school.
$y=ax^2+bx+c$
where $a,b,c$ must all be integers. (i.e. no score)

example
$y=x^2-3x+2$
untie
1 -1
1 -2
The two 1s in the first column are from $X^2$ ($x^2=x x$)
The 1 and 2 in the second column are from $c$, in this question $c=2$, 2 can be split into $1 2$ or $-1*-2$

The logic of splitting: make $b$ by splitting $c$ into the form of $c1*c2$
See below for details

After the cross is connected, the next step is to multiply the 1 and -2 of the first row to get $1*-2=-2$
Multiply the 1 in the second row by -1 to get $1*-1=-1$
Then add the two results $-2+(-1)=-3=b$
$b$ has been figured out. At this step, the cross multiplication is right. Finally, the simplified formula can be written horizontally and the 1 in the first column of the formula can be multiplied by x again.
1 plus -1 in the first row times 1 plus -2 in the second row
$(1 x)+-1 (1 x)+-2=(x-1) (x-2)$
Finish

$y=ax^2+bx+c$

On the coordinate axis $y=ax^2+bx+c$ is a U-shaped curve, either the opening is facing up or the opening is facing down.
The ± of $a$ determines the opening direction + upward – downward.

$\Delta=b^2-4ac$
$\Delta$ is the position of the midpoint of the curve on the Y axis
The $\Delta>0$ function has two points on the X axis (y=0) $x_1,x_2$
The $\Delta=0$ function has a point on the X axis (y=0) $x_1=x_2$
The $\Delta<0$ function has no point on the X axis (y=0) and does not intersect the X axis

Where did $\Delta$ come from?
$b^2-4ac$ is derived from the general formula $ax^2+bx+c=0$:
The specific derivation process of $b^2-4ac$:
$ax^2+bx+c=0(a≠0)$
Divide both sides by a
get $x^2+b/aX+c/a=0$
Re-formulation is $x^2+b/aX+(b/2a)^2=-c/a+(b/2a)^2$
$x+b/2a)^2=b²-4ac/4a^2$
If $b²-4ac$ is greater than or equal to 0
$x=-b$±$b^2-4ac/2a$ under the square root

The symmetry axis of the function curve is $x=-\frac{b}{2a}$

### Unary quadratic function universal formula

It’s called Mitternacht Formel in German
$x_1,_2=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

### Veda’s Theorem

Use it when it’s hard to come by.
Premise: $\Delta>0$ (with $x_1,x_2$)
$x_1+x_2=-\frac{b}{a}$
$x_1x_2=\frac{c}{a}$

The ± of $a$ determines the opening direction + upward – downward.