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The solution of quadratic equations

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This article considers the standard quadratic equation of the form:

The article derives a formula for the roots of a quadratic equation by the method of complementing to a full square, numerical values ​​instead a, b, c will not be substituted.

ax 2 + bx + c = 0 2 Divide both sides of the equation by but.

x 2 + (b / a) x + c / a = 0 3 Subtract s / a from both sides of the equation.

x 2 + (b / a) x = -c / a 4 Divide the coefficient at x (b / a) by 2, and then square the result. Add the result to both sides of the equation.

x 2 + (b / a) x + b 2 / 4a 2 = -c / a + b 2 / 4a 2 5 Simplify the expression by factoring the left side and adding the terms on the right side (first find the common denominator).

(x + b / 2a) (x + b / 2a) = (-4ac / 4a 2) + (b 2 / 4a 2)

(x + b / 2a) 2 = (b 2 - 4ac) / 4a 2 6 Extract the square root from each side of the equation.

√ ((x + b / 2a) 2) = ± √ ((b 2 - 4ac) / 4a 2)

x + b / 2a = ± √ (b 2 - 4ac) / 2a 7 Subtract b / 2a from both sides and you get the formula for the roots of the quadratic equation.

Discriminant

Let the quadratic equation ax 2 + bx + c = 0 be given. Then - this is just the number D = b 2 - 4 ac.

This formula must be known by heart. Where it comes from is now unimportant. Another thing is important: by the sign of the discriminant you can determine how many roots the quadratic equation has. Namely:

  1. If D D = 0, there is exactly one root,
  2. If D> 0, there will be two roots.

Note: the discriminant indicates the number of roots, and not at all their signs, as for some reason many believe. Take a look at the examples and you will understand everything yourself:

Task. How many roots have quadratic equations:

  1. x 2 - 8 x + 12 = 0,
  2. 5 x 2 + 3 x + 7 = 0,
  3. x 2 - 6 x + 9 = 0.

We write out the coefficients for the first equation and find the discriminant:
a = 1, b = −8, c = 12,
D = (−8) 2 - 4 · 1 · 12 = 64 - 48 = 16

So, the discriminant is positive, so the equation has two different roots. Similarly, we analyze the second equation:
a = 5, b = 3, c = 7,
D = 3 2 - 4 · 5 · 7 = 9 - 140 = −131.

The discriminant is negative, there are no roots. The last equation remains:
a = 1, b = −6, c = 9,
D = (−6) 2 - 4 · 1 · 9 = 36 - 36 = 0.

The discriminant is zero - the root will be one.

Note that coefficients were written for each equation. Yes, it’s a long time, yes, it’s boring - but you won’t mistake the coefficients and make silly mistakes. Choose for yourself: speed or quality.

By the way, if you “get your hand in it,” after a while you’ll no longer need to write out all the odds. You will perform such operations in your head. Most people begin to do so somewhere after 50-70 solved equations - in general, not so much.

The roots of the quadratic equation

Now let's move on to the solution. If the discriminant is D> 0, the roots can be found by the formulas:

The basic formula of the roots of the quadratic equation

When D = 0, you can use any of these formulas - you get the same number, which will be the answer. Finally, if D x 2 - 2 x - 3 = 0,

  • 15 - 2 x - x 2 = 0,
  • x 2 + 12 x + 36 = 0.
  • First equation:
    x 2 - 2 x - 3 = 0 ⇒ a = 1, b = −2, c = −3,
    D = (−2) 2 - 4 · 1 · (−3) = 16.

    D> 0 ⇒ the equation has two roots. Find them:

    The second equation:
    15 - 2 x - x 2 = 0 ⇒ a = −1, b = −2, c = 15,
    D = (−2) 2 - 4 · (−1) · 15 = 64.

    D> 0 ⇒ the equation again has two roots. Find them

    Finally, the third equation:
    x 2 + 12 x + 36 = 0 ⇒ a = 1, b = 12, c = 36,
    D = 12 2 - 4 · 1 · 36 = 0.

    D = 0 ⇒ the equation has one root. You can use any formula. For example, the first:

    As you can see from the examples, everything is very simple. If you know the formulas and be able to count, there will be no problems. Most often, errors occur when substituting negative coefficients in the formula. Here again, the technique described above will help: look at the formula literally, write down each step - and very soon get rid of errors.

    Incomplete quadratic equations

    It happens that the quadratic equation is somewhat different from what is given in the definition. For example:

    It is easy to notice that one of the terms is absent in these equations. Such quadratic equations are even easier to solve than the standard ones: they do not even need to consider the discriminant. So, we introduce a new concept:

    The equation ax 2 + bx + c = 0 is called if b = 0 or c = 0, i.e. the coefficient of the variable x or the free element is zero.

    Of course, a very difficult case is possible when both of these coefficients are equal to zero: b = c = 0. In this case, the equation takes the form a x 2 = 0. Obviously, this equation has a single root: x = 0.

    Consider the remaining cases. Let b = 0, then we get an incomplete quadratic equation of the form ax 2 + c = 0. We slightly transform it:

    Solution of an incomplete quadratic equation

    Since the arithmetic square root exists only from a non-negative number, the last equality only makes sense for (- c / a) ≥ 0. Conclusion:

    1. If the inequality (- c / a) ≥ 0 holds in an incomplete quadratic equation of the form ax 2 + c = 0, there will be two roots. The formula is given above
    2. If (- c / a) c / a) ≥ 0. It is enough to express the quantity x 2 and see what is on the other side of the equal sign. If there is a positive number, there will be two roots. If negative, there will be no roots at all.

    Now we will deal with equations of the form ax 2 + bx = 0, in which the free element is equal to zero. Everything is simple here: there will always be two roots. It is enough to factor the polynomial:

    Bracketing the common factor

    The product is zero when at least one of the factors is zero. From here are the roots. In conclusion, we analyze several such equations:

    Task. Solve quadratic equations:

    x 2 - 7 x = 0 ⇒ x · (x - 7) = 0 ⇒ x 1 = 0, x 2 = −(−7)/1 = 7.

    5 x 2 + 30 = 0 ⇒ 5 x 2 = −30 ⇒ x 2 = −6. There are no roots, because a square cannot be equal to a negative number.

    4 x 2 - 9 = 0 ⇒ 4 x 2 = 9 ⇒ x 2 = 9/4 ⇒ x 1 = 3/2 = 1,5, x 2 = −1,5.

    Examples of quadratic equations

    • 5x 2 - 14x + 17 = 0
    • −x 2 + x +
      1
      3
      = 0
    • x 2 + 0.25x = 0
    • x 2 - 8 = 0

    To find "a", "b" and "c" you need to compare your equation with the general form of the quadratic equation "ax 2 + bx + c = 0".

    Let's practice defining the coefficients a, b, and c in quadratic equations.

    The equationOdds
    5x 2 - 14x + 17 = 0
    • a = 5
    • b = −14
    • c = 17
    −7x 2 - 13x + 8 = 0
    • a = −7
    • b = −13
    • c = 8
    −x 2 + x +
    1
    3
    = 0
    • a = −1
    • b = 1
    • c =
      1
      3
    x 2 + 0.25x = 0
    • a = 1
    • b = 0.25
    • c = 0
    x 2 - 8 = 0
    • a = 1
    • b = 0
    • c = −8

    How to solve quadratic equations

    In contrast to linear equations, a special formula for finding the roots is used to solve quadratic equations.

    To solve the quadratic equation you need:

    • reduce the quadratic equation to the general form "ax 2 + bx + c = 0". That is, only “0” should remain on the right side,
    • use the formula for the roots:

    x1,2 =
    −b ± √ b 2 - 4ac
    2a

    Let's look at an example of how to apply the formula to find the roots of a quadratic equation. Solve the quadratic equation.

    The equation "x 2 - 3x - 4 = 0" has already been reduced to the general form "ax 2 + bx + c = 0" and does not require additional simplifications. To solve it, we just need to apply the formula for finding the roots of the quadratic equation.

    Define the coefficients "a", "b" and "c" for this equation.

    The equationOdds
    x 2 - 3x - 4 = 0
    • a = 1
    • b = −3
    • c = −4

    Substitute them in the formula and find the roots.

    x 2 - 3x - 4 = 0
    x1,2 =
    −b ± √ b 2 - 4ac
    2a

    x1,2 =
    −(−3) ± √ (−3) 2 − 4 · 1· (−4)
    2 · 1

    x1,2 =
    3 ± √ 9 + 16
    2

    x1,2 =
    3 ± √ 25
    2

    x1,2 =
    3 ± 5
    2

    x1 =
    3 + 5
    2
    x2 =
    3 − 5
    2
    x1 =
    8
    2
    x2 =
    −2
    2
    x1 = 4x2 = −1

    Answer: x1 = 4, x2 = −1

    Be sure to memorize the formula for finding the roots.

    x1,2 =
    −b ± √ b 2 - 4ac
    2a

    With its help, any quadratic equation is solved.

    In the formula "x1,2 =
    −b ± √ b 2 - 4ac
    2a
    »Often replace radical expression
    "B 2 - 4ac" in the letter "D" and is called the discriminant. The concept of discriminant is discussed in more detail in the lesson “What is discriminant”.

    Consider another example of a quadratic equation.

    In this form, determining the coefficients "a", "b" and "c" is rather difficult. Let's first bring the equation to the general form "ax 2 + bx + c = 0".

    Now you can use the formula for the roots.

    x1,2 =
    −(−6) ± √ (−6) 2 − 4 · 1 · 9
    2 · 1

    x1,2 =
    6 ± √ 36 − 36
    2

    x1,2 =
    6 ± √ 0
    2

    x1,2 =
    6 ± 0
    2

    x =
    6
    2

    x = 3
    Answer: x = 3

    There are times when there are no roots in quadratic equations. This situation arises when a negative number appears in the formula under the root.

    We remember from the definition of the square root that it is impossible to extract the square root from a negative number.

    Consider an example of a quadratic equation that has no roots.

    5x 2 + 2x = - 3
    5x 2 + 2x + 3 = 0
    x1,2 =
    −2 ± √ 2 2 − 4 · 3 · 5
    2 · 5

    x1,2 =
    −2 ± √ 4 − 60
    10

    x1,2 =
    −2 ± √ −56
    10

    Answer: there are no valid roots.

    So, we got a situation where a negative number is under the root. This means that there are no roots in the equation. Therefore, in response, we wrote “No real roots”.

    What do the words “no real roots” mean? Why can't you just write “no roots”?

    In fact, there are roots in such cases, but they do not go through the school curriculum, therefore, in response, we record that among the real numbers there are no roots. In other words, "There are no real roots."

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