Have you forgotten how to solve an incomplete quadratic equation?

2024 Author: Angel Austin | [email protected]. Last modified: 2024-01-09 16:27

How to solve an incomplete quadratic equation? It is known that it is a particular version of the equality will be zero - simultaneously or separately. For example, c=o, v ≠ o or vice versa. We almost remembered the definition of a quadratic equation.

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The trinomial of the second degree is equal to zero. Its first coefficient a ≠ o, b and c can take on any values. The value of the variable x will then be the root of the equation when, when substituting, it turns it into the correct numerical equality. Let us dwell on real roots, although complex numbers can also be solutions to the equation. It is customary to call an equation complete if none of the coefficients is equal to o, but ≠ o, to ≠ o, c ≠ o.

Solve an example. 2x²-9x-5=oh, we find

D=81+40=121, D is positive, so there are roots, x ₁ =(9+√121):4=5 and the second x_{2=(9-√121):4=-o, 5. Checking will help make sure they are correct.}

Here is a step-by-step solution to the quadratic equation

Through the discriminant, you can solve any equation, on the left side of which there is a known square trinomial with a ≠ o. In our example. 2x²-9x-5=0 (ax^2+in+s=o)

• First, find the discriminant D using the known formula in2-4ac.
• Checking what the value of D will be: we have more than zero, it can be equal to zero or less.

• We know that if D › o, the quadratic equation has only 2 different real roots, they are denoted x₁ usually and x₂, this is how it was calculated:

  x₁=(-v+√D):(2a), and the second: x_{2=(-in-√D):(2a).}

• D=o - one root, or, they say, two equal:

  x₁ equal to x_{2 and equals -v:(2a).}

• Finally, D ‹ o means that the equation has no real roots.

Let's consider what are incomplete equations of the second degree

1. ax²+in=o. The free term, the coefficient c at x⁰, is zero here, at ≠ o.

   How to solve an incomplete quadratic equation of this kind? Let's take x out of brackets. Remember when the product of two factors is zero.

   x(ax+b)=o, this can be when x=o or when ax+b=o.

   Solving the 2nd linear equation;

   x₂ =-b/a.

2. Now the coefficient of x is o and c is not equal (≠)o.

   x²+s=o. Let's move from to the right side of the equality, we get x² =-с. This equation only has real roots when -c is a positive number (c ‹ o), x₁ then equals √(-c), respectively x _{2 ― -√(-s). Otherwise, the equation has no roots at all.}

3. Last option: b=c=o, i.e. ah2=o. Naturally, such a simple equation has one root, x=o.

Special cases

How to solve an incomplete quadratic equation was considered, and now we will take any kind.

In the full quadratic equation, the second coefficient of x is an even number.

Let k=o, 5b. We have formulas for calculating the discriminant and roots.

D/4=k²-ac, the roots are calculated like this x_{1, 2} =(-k±√(D/4))/a for D › o.

x=-k/a for D=o. No roots for D ‹ o.

There are reduced quadratic equations, when the coefficient of x squared is 1, they are usually written x² +px+ q=o. All of the above formulas apply to them, but the calculations are somewhat simpler. +9, D=13.

x¹ =2+√13, x² =2-√13.

Besides, Vieta's theorem can be easily applied to the given ones. It says that the sum of the roots of the equation is -p, the second coefficient with a minus (meaning the opposite sign), and the product of these same roots will be equal to q, the free term. Check out howit would be easy to verbally determine the roots of this equation. For non-reduced (for all non-zero coefficients) this theorem is applicable as follows: 1_x2 _{equal to/a.}

The sum of the free term c and the first coefficient a is equal to the coefficient b. In this situation, the equation has at least one root (it is easy to prove), the first one is necessarily equal to -1, and the second - c / a, if it exists. How to solve an incomplete quadratic equation, you can check it yourself. As easy as pie. Coefficients can be in some ratios among themselves

• x²+x=o, 7x^2-7=o.

• The sum of all coefficients is o.

  The roots of such an equation are 1 and c/a. Example, 2x²-15x+13=o.

  x₁ =1, x_2=13/2.

There are a number of other ways to solve different equations of the second degree. Here, for example, is a method for extracting a full square from a given polynomial. There are several graphic ways. When you often deal with such examples, you will learn to "click" them like seeds, because all the ways automatically come to mind.