Equation theory

General rule:

What I do on the left side, I also have to do on the right side.

 A + B = C + D

F is added + F:

A + B + F = C + D + F  

Equations with an unknown x, the value of x is sought.

ax + b = c

The equation is simplified so far that only the unknown (x) on the left or on the right side remains :  

So we solve for x: Apply the rule:

Subtract b then divide by a

ax + b – b = c – b and ax = c – b,

division by a yields

ax / a = c – b / a

x = c – b / a

Control: we now use this result of x in ax + b = c:

a (c – b) / a + b = c and get: a disappears, leaving

c- b + b = c  

Further examples of equations with 1 unknown

 

1 / y = 1/2, searched y:

we reverse the fraction and get

y = 2

control ½ = ½  

 

1 / z = z / 2, searched z:

Inverse fraction returns z = 2 / z then multiply by z:

z² = 2 then

z = √2,

Control 1 / √2 = √2 / 2 or 2exp-½ = 2exp (½ – 1)

 

x² = a,

x = √a

control √a * √a = a  

 

1 / y = y – 1 multiply by y yields:

y² – y = 1

Now we already have a 2nd order equation or a quadratic equation:

this has the form ax² + bx + c = 0

y² – y – 1 = 0

The solution is the so-called Discriminant, D

D = b² – 4ac If D> 0 (D is greater than 0) there are 2 solutions

x (1) = -b + √D / 2a,

x (2) = -b – √D / 2a

If D = 0, there is 1 solution

x (1) = x (2) = – b / 2a

if D <0 there are no solutions

Back to example: y² – y – 1 = 0

D = 1 + 4 = 5,

y1 = (1 + √5) / 2,

y2 = (1 – √5) / 2

So this equation has 2 solutions

 

Now about the Discriminant, where does it come from?

Derivation of the Discriminant, D:

ax² + bx + c = 0 Almost every field of study or study comes into contact with this equation. Be it at the Matura, Abi, vocational baccalaureate, technical college of technology and economics or university scientific, technical or economic direction. The solution of this equation is simply simple, if one knows at least the discriminant: D = b² – 4ac. If D is positive and> 0 we get 2 solutions if D = 0, 1 solution. If D <0, the equation for real numbers, R, contains no solution. But do not memorize the relatively simple formula for one time.

 Simply derive D by using numbers in the equation ax² + bx + c = 0, which provide only a single solution for x. In the last chapter (Sings, symbols and relationships) you will find the binomial formula, (a + b) ² under “Important Relations”.

It gives the decomposition (a + b)² =  a² + 2ab + b²

For this equation, ax² + bx + c = 0 we now look for some whose x1 = x2.

We now decompose a binomial formula (2 + 1) ² = 4 + 4 + 1.

Now we put this result in the quadratic equation instead of a, b and c:

4x² + 4x + 1 = 0,

this equation has only 1 solution: x1 = x2 = -1/2.

we calculate 4 * (- 0.5) ² + 4 (-0.5)  + 1 = 1 – 2 + 1 = 0

This result x = -1/2 also satisfies the equation 2x + 1 = 0, which is why 4x² + 4x + 1 = 0 has only one solution x1 = x2 = -1/2.

Generally one can write: -1/2 = -b / 2a, because -4 / 2 * 4 = -1/2.

We now determine the discriminant D: As we know, the quadratic equation only has 2 different solutions if D> 0. A root equation of the format x² + a = 0

If the solution is √a, if a = 4, then x1 = 2 and x2 = -2. So the discriminant D must be the sought root. In the quadratic equation ax² + bx + c = 0,  x is now replaced by the presumed solution of the equation

x = -b / 2a + √D / 2a and obtain the following:

a * {(- b -√D) / 2a} ² + b (-b -√D) / 2a + c = 0,

multiply, shorten with a and make the denominator the same:

(b² + 2b√D + D -2b² – 2b√D + 4ac) / 4a = 0,

simplify and make roots disappear

– b² + D + 4ac = 0,

and now solve for D

D = b² – 4ac

 

Excel formula for solving the quadratic equation: ax² + bx + c = 0,

a = A1, b = B1, c = C1

Enter the real numbers a, b and c in any empty cell

Discriminant = B1 * B1-4 * A1 * C1

x1 = (-B1 + ROOT (B1 * B1-4 * A1 *C1))/2*A1 

x2 = -B1 + ROOT (B1 * B1-4 * A1 * C1)/2 *A1  

 

Equation system with 2 unknowns

a11x1 + a12x2 = b1

a21x1 + a22x2 = b2

 

first solve for x1:

x1 = (b1 – a12x2) / a11

fill in this result the 2nd equation and solve for x2

x2 = (b2a11 – a21b1) / (a22a11 – a21a12)

We use this value again in the first equation and solve for x1.

results:

x1 = (a22b1-a12b2) / (a11a22-a21a12)

x2 = (a11b2 – a12b1) / (a11a22 – a21a12)

EXERCISE: Check by substituting the solutions of x1 and x2 in the above equation with 2 unknowns.

 

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Equation with 2 unknowns, solution with Determinants:

 

Solution with Excel program:

 

More equations • equations of third degree

ax³ + bx² + cx + d = 0

 

 

Differential Equations

f ‘(x) = f (x) – a f ” (x) = -Da

In these equations, the unknown sought is a particular function f (x) •

Equation system with 3 and more unknowns

a11x1 + a12x2 + a13x3 = b1

a21x1 + a22x2 + a23x3 = b2

a31x1 + a32x2 + a33x3 = b3

Approaches are made using the determinant method Solutions below and in higher lessons.

 

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