Square equations - the base on which almost all school mathematics is built. But it happens that the foundations are flying out of the head. In this article we will examine the types of square equations in detail and solve them, so you can easily solve them yourself.
What is square equations?
This is the view equations aX.2 + bX. + c. = 0
where, a ≠ 0, b, c - numbers; X - variable.
Equations are without roots, with one root and two different roots.
Find the roots in two ways:
- through discriminant;
- on the Vieta Theorem.
Discriminant
We find it according to the formula d \u003d b 2 - 4ac.
Actually, according to the resulting answer and determine:
- D \u003c0, no roots;
- D \u003d 0, only one root;
- D\u003e 0, two roots.
We find roots by formulas:
1. No roots.
2. X \u003d -B / 2A
3. x1 \u003d (-b + √D) / 2a; x2 \u003d (-b - √D) / 2A.
Example:
1. 3x 2 + 4X + 3 \u003d 0
a \u003d 3; b \u003d 4; c \u003d 3;
D \u003d 4. 2 - 4 · 3 · 3 \u003d 0.
No roots.
2. X. 2 - 6x + 9 \u003d 0.
a \u003d 1; b \u003d -6; C \u003d 9;
D \u003d (-6) 2 - 4 · 1 · 9 \u003d 36 - 36 \u003d 0.
x \u003d -b / 2a \u003d 6/2 \u003d 3
One root: x \u003d 3
3. X. 2 - 5x + 6 \u003d 0
a \u003d 1; b \u003d -5; C \u003d 6;
D. = b.2 - 4. aC \u003d (-5) 2.- 4 · 1 · 6 \u003d 25 - 24 \u003d 1
x1 \u003d ( −(−
+√1) / 2·1 = 3x2 \u003d ( −(−
−√1) / 2·1 = 2Answer: x1 \u003d 3; x2 \u003d 2.
Vieta theorem
The reduced square equation of the form:
- x. 2 + px + Q \u003d 0
Coefficient a \u003d 1, the amount of the roots \u003d −p, work \u003d Q.
If x1 and x2 are the roots of the given square equation, then:
x. 2 + px + Q \u003d 0
x1 + x2 \u003d −p; x1. · x2 \u003d Q.
Theorem, reverse theorem of Vieta
If p, q, x1, x2 are such that:
x1 + x2 \u003d −p; x1. · x2 \u003d Q.
then x1, x2 - roots equation x 2 + px + Q \u003d 0
Example:
x. 2 - 10x + 21 \u003d 0.
x1 + x2 \u003d 10; x1. · x2 \u003d 21.
It is easy to notice that these equalities are suitable for numbers 3 and 7.
Exceptions
But in solving equations there are special cases - incomplete equations.
- a. x.2+ C \u003d 0, B equals 0;
- a. x.2 + BX \u003d 0, C is 0;
- a. x.2 \u003d 0, b and C are 0.
But you should not worry: such equations are easily solved (you can solve through discriminant).
Example:
5x.2 = 0
5x.2/ 5 \u003d 0/5
x.2 = 0
x. = 0
Answer: x. = 0
That's all! As you can see, it was not so difficult to solve the square equations, so now it's about you.
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