Quadratic equations

Finding roots

Because the highest power in the equation Ax² + Bx + C = 0 is x², it is called an equation of degree 2 or a quadratic equation. In what follows, we will assume that A 0. If B = 0, you have a purely quadratic, if B 0, a mixed quadratic equation.

I. B = 0:

Task: Find the roots of x² = 9.

(+3)² as well as (-3)² yield 9. Hence there are two roots: x₁ = +3, x₂ = -3. Since we understand by only the principal value, we must add the sign, whence x² = a has the roots x₁=+, x₂= -.

II. C = 0:

Task: Find the roots of x² - 8x = x(x - 8) = 0 . You have either x = 0 or x - 8 = 0, whence x₁ = 0, x₂ = 8.

III. B 0, C 0:

Task: Find the roots of x² + 6x + 8 = 0. You compare the left hand side with a² + 2ab + b² = (a + b)² and recognize that b must have the value 3 and therefore b² = 9; in order to complete the square, you must add 9 to both sides of the equation x² + 6x = - 8, whence (x+3)² = 1 or x + 3 = 1, that is x₁ = -2, x₂=-4. Remember that

the quadratic complement is the square of half the coefficient of x!

General solution of the equation x² + ax + b = 0 (standard form)

Add on both sides of the equation x² + ax = - b the quadratic complement (a/2)² to obtain

x²+ax+(a/2)²=(a/2)²-b

(x+a/2)²=(a/2)²-b

Task 1: Find the roots of x² -10x + 16 = 0.

x = 5 3

x1 = 8

x2 = 2

Task 2: Find the roots of 4x² + 8x - 28 - 6x = 7x² + 23x + 8.

Combine the quadratic, linear and constant terms to obtain:

x²+7x+12=0 = 0				x=-7/22
		x1= -3		x2= -4

If you want to solve the general form of the quadratic equation Ax²+Bx+C=0, you first rewrite it in the form x² + Bx/A + C/A = 0 and then find for the roots the formulae

or

Discriminant

Whether a quadratic equation has real roots will depend on whether the term under the root in

is positive or negative. That is why the term discriminant D = a² - 4b has been introduced. If D > 0, you have two real roots, if D = 0, you have double roots. A difficulty arises if D < 0. For example, the equation x²+2x+5=0 is such a case, because

There does not exist a real number the square of which is -4; you will, of course, suspect immediately that mathematics has removed this limitation. Such roots are called complex roots, whence you have the Rule:

A quadratic equation has two solutions which are either real, coincide or are complex.

In terms of the general equation, you then have D = B² - 4AC.

Vieta's theorem

Since it is clear that the roots of an equation depend on its constants, you will suspect that there exist certain relations between the constants and the roots. If you examine the three equations

you will note, for example, that 15 + 2 = 17 and 15·2 = 30 and the same is true for the other two equations. The general proof for this result follows:

The roots of x² + ax + b = 0 are

,

whence x₁ + x₂ = -a and

These relations were discovered by Vieta (1540 - 1603), whence they are called Vieta's Theorem:

x₁ + x₂ = -a, x₁ · x₂ = b.

The sum of the roots of a quadratic equation x² + ax + b = 0 is equal to the negative coefficient a of x, their product to the constant term.

You can use this rule to check you solutions as well as to set up quadratic equations with predetermined roots.

Task: Construct the quadratic equation with the roots +1 and +3.

The required equation is x² - 4x + 3 = 0.

Every expression of the form x² + ax + b can be transformed into a product. If you substitute a=-(x₁ + x₂) and b = x₁·x₂ , it becomes

x² - (x₁+ x₂) + x₁·x₂ = x² - x₁·x - x₂x + x₁·x₂ =
=x(x - x₁) - x₂(x - x₁) = (x - x₁)(x - x₂),

whence

x² + ax + b = (x - x₁)(x - x₂).

Task: Simplify (x² - 6x + 8)/(x - 4).

The roots of the quadratic equation x² - 6x + 8 are 2 and 4, whence

(x² -6x + 8)/(x - 4) = (x - 2)(x - 4)/(x - 4) = x - 2.

This can always be done except when x = 4, because you may not divide by 0.

Exercises: Find the roots of the quadratic equations:

Answer

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