If in the quadratic equation *ax*^{2 }+ *bx *+ *c *= 0 the second factor *b* is even, then the solution of this equation can be simplified a little. The discriminant for such an equation can be calculated by the formula *D*_{1 }= *k*^{2 }β *ac*, and the roots by the formulas and .

## Examples

Solve the quadratic equation *x*^{2 }+ 6*x *β 16 = 0. In this equation, the second coefficient is even. To use the formulas for the even coefficient, we first need to know what the variable *k* is equal to.

Any even number n can be represented as the product of number 2 and number *k*, that is, 2*k*.

*n* = 2*k*

For example, the number 10 can be represented as 2 Γ 5.

10 = 2 Γ 5

In this product *k* = 5.

The number 12 can be represented as 2 Γ 6.

12 = 2 Γ 6

In this product *k* = 6.

The number -14 can be represented as 2 Γ (-7)

In this product *k* = -7.

As we see, the factor 2 does not change. Only the factor k changes.

In the equation *x*^{2 }+ 6*x β *16 = 0, the second coefficient is 6. This number can be represented as 2 Γ 3. In this product *k* = 3. Now we can use the formulas for the even coefficient.

Find the discriminant by the formula* D*_{1 }= *k*^{2 }β *ac*

*D*_{1 }= *k*^{2 }β *ac* = 3^{2 }β 1 Γ (β16) = 9 + 16 = 25

Now let's calculate the roots using the formulas: and .

So the roots of the equation x^{2} + 6x - 16 = 0 are the numbers 2 and -8.

Unlike the standard formula for calculating the discriminant *(D=b ^{2} - 4ac)*, the formula

*D1 = k*does not need to multiply the number 4 by

^{2}- ac*ac*.

And unlike formulas and , formulas and do not contain the factor 2 in the denominator, which again frees us from additional calculations.

**Example 2.** Solve the quadratic equation 5*x*^{2 }β 6*x *+ 1=0

The second coefficient is an even number. It can be represented as 2 Γ (-3). That is, *k = -3*. Find the discriminant by the formula *D*_{1 }= *k*^{2 }β *ac*

*D*_{1 }= *k*^{2 }β *ac = *(β3)^{2 }β 5 Γ 1 = 9 β 5 = 4

The discriminant is greater than zero. So the equation has two roots. To calculate them we use formulas and

**Example 3.** Solve the quadratic equation *x*^{2 }β 10*x *β 24 = 0

The second coefficient is an even number. It can be represented as 2 Γ (-5). That is, k = -5. Find the discriminant by the formula *D*_{1 }= *k*^{2 }β *ac*

*D*_{1 }= *k*^{2 }β *ac = *(β5)^{2 }β 1 Γ (β24) = 25 + 24 = 49

The discriminant is greater than zero. So the equation has two roots. To calculate them we use formulas and

The usual way to determine the number k is to divide the second coefficient by 2.

Indeed, if the second quotient *b* is an even number, it can be represented as *b = 2k*. In order to express the factor k from this equality, one must divide the product of *b* by the factor 2

For example, in the previous example you could simply divide the second factor -10 by 2 to determine the number *k*

**Example 4.** Solve a quadratic equation

The coefficient of *b* is . This expression consists of the multiplier 2 and the expression . That is, it is already represented as 2*k*. It turns out that

Find the discriminant by the formula *D*_{1 }= *k*^{2 }β *ac*

The discriminant is greater than zero. So the equation has two roots. To calculate them we use formulas and

Calculating the root of the equation produces a fraction that contains the square root of 2. The square root of 2 can only be extracted approximately. If you do this approximate extraction and then add the result to 2, and then divide the numerator by the denominator, you get a not very nice answer.

In such cases, the answer is written down without performing approximate calculations. In our case, the first root of the equation is .

Calculate the second root of the equation:

## How to derive formulas

Let's visualize how formulas for calculating the roots of a quadratic equation with an even second coefficient appeared.

Consider the quadratic equation *ax*^{2 }+ *bx *+ *c *= 0. Assume that the coefficient of b is an even number. Then it can be denoted as 2*k*

*b = *2*k*

Replace in the equation *ax*^{2 }+ *bx *+ *c *= 0 the quotient of *b* by the expression 2*k*

*ax*^{2 }+ 2*kx *+ *c *= 0

Now let's calculate the discriminant using the previously known formula:

*D = b*^{2 }*β *4*ac = *(2*k*)^{2 }*β *4*ac = *4*k*^{2 }*β *4*ac*

In the resulting expression, take the common factor 4 out of the brackets.

*D = b*^{2 }*β *4*ac = *(2*k*)^{2 }*β *4*ac = *4*k*^{2 }*β *4*ac *= 4(*k*^{2 }β *ac*)

What can be said about the resulting discriminant? If the second coefficient is even, it consists of factor 4 and the expression *k*^{2 }β *ac*.

In the expression 4(*k*^{2 }β *ac*) the factor 4 is constant. So the sign of the discriminant depends on the expression *k*^{2 }β *ac*. If this expression is less than zero, then *D* will be less than zero. If the expression is greater than zero, then *D* is greater than zero. If the expression is equal to zero, then *D* will be equal to zero.

That is, the expression *k*^{2 }β *ac* is the discriminant. Such a discriminant is usually denoted by the letter *D*_{1}

*D*_{1 }= *k*^{2 }β *ac*

Now let's see how formulas and are derived

In our equation *ax*^{2 }+ *bx *+ *c *= 0 the quotient of *b* is replaced by the expression 2*k*. Use the standard formulas for calculating the roots. That is, we use formulas and . Only instead of b we substitute 2*k*. Also note that D equals the expression 4(*k*^{2 }β *ac*).

But it was said earlier that the expression *k*^{2 }β *ac* is denoted by *D*_{1}. Then in our transformations we should make this substitution as well:

Now let's calculate the square root located in the numerator. This is the square root of the product - it is equal to the product of the roots. Let's rewrite the rest without changing:

Now, in the resulting expression, take the common factor 2 out of the brackets.

Let's reduce the resulting fraction by 2

Similarly, the formula for calculating the second root is derived:

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