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In mathematical form, the term second-degree equation can be defined as:

Quadratic equations formula in general form is given as:

Where,

- “a” and “b” represents the leading coefficient
- “c” represents the constant number which is the absolute term of the equation

- 3x^2 + x + 5 = 0,
- -x^2 + 7x + 5 = 0,
- x^2 + x = 0.

When a quadratic polynomial is kept equal to zero, it becomes a quadratic equation. The values of x that satisfy the equation are known as the roots of the squared equation. The second degree equation will always have two nature roots that may be either real or imaginary. Whether you are trying to resolve problems regarding quadratic expression, you can use this Quadratic formula calculator online quickly.

The solution of quadratic equation roots is solved by the quadratic formula:

1. The roots of the second degree equation are given by:

- x = (-b ± √D)/2a,

where D = b2 – 4ac

2. Sum and Product of roots:

If α and β are the quadratic equation roots, then we can say that

- Sum of roots = α+β= -b/a = coefficient of x/coefficient of x^2
- Product of roots = αβ = c/a = constant term/coefficient of x^2

3. Quadratic equation in the form of roots:

- x^2 – (α+β)x + (αβ) = 0

4. The quadratic equations a1x2 + b1x + c1 = 0 and a2x2 + b2x + c2 = 0 have;

- One common quadratic square root if (b1c2 – b2c1)/(c1a2 – c2a1) = (c1a2 – c2a1)/(a1b2 – a2b1)
- Both roots common if a1/a2 = b1/b2 = c1/c2

5. In quadratic equation ax^2 + bx + c = 0 or [(x + b/2a)^2 – D/4a^2]

- If a > 0, minimum value = 4ac – b^2/4a at x = -b/2a.
- If a < 0, maximum value 4ac – b^2/4a at x= -b/2a.

6. If α, β, γ are roots of cubic equation ax^3 + bx^2 + cx + d = 0, then, α + β + γ = -b/a, αβ + βγ + λα = c/a, and αβγ = -d/a

7. A second degree equation becomes an identity (a, b, c = 0) if the polynomial equation is satisfied by more than two numbers i.e. having more than two roots or solutions that are real or complex.

Besides that, you can also calculate the values of squared equation roots by using an online Synthetic Division Calculator swiftly.

The root of a quadratic equation can be defined as:

In other words, x = α is a root of the second degree equation f(x), if f(α) = 0.

The real roots of the second degree equation f(x) = 0 are the x-coordinates of the points where the curve y = f(x) intersect the x-axis.

- If c = 0 then one of the quadratic equation roots is zero and the other is -b/a
- If b = c = 0 then both the roots are zero
- If a = c the roots are reciprocal to each other

Discriminant of a quadratic equation can be represented as:

The discriminant of a squared equation indicates the nature of roots.

Nature of Roots of Quadratic Equation

- If the value of discriminant = 0 i.e. b^2 – 4ac = 0

The roots of the quadratic equation will have equal i.e. α = β = -b/2a

- If the value of discriminant < 0 i.e. b^2 – 4ac < 0

The roots of the quadratic equation will have imaginary i.e. α = (p + iq) and β = (p – iq). Where ‘iq’ in the equation is the imaginary part of a complex number

- If the value of discriminant (D) > 0 i.e. b^2 – 4ac > 0

The roots of the quadratic equation will have real

- If the value of discriminant (D) > 0 and D is a perfect square

The roots of the quadratic equation will have rationals

- If the value of discriminant (D) > 0 and but D is not a perfect square

The roots of the quadratic equation will have irrationals i.e. α = (p + √q) and β=(p – √q)

- If the value of discriminant > 0, D is a perfect square, a = 1, and b and c are integers

The roots of the quadratic equation will have an integral.

However, you can also determine the zeros or roots of a quadratic expression to find the solution of an equation by using an online Zeros calculator within a few clicks.

Find the second degree equation having the roots 6 and 9 respectively.

The quadratic formula equation having the roots α, β, is x^2 - (α + β)x + αβ = 0.

Given α = 6, and β = 9.

Therefore the quadratic expression is:

x^2 - (6 + 9)x + 6×9 = 0

x^2 - 15x + 54 = 0

Answer: Hence the required quaratic equation is x^2 - 15x + 54 = 0

Find the second degree equation with rational coefficients when one root is 1/(2 + √5).

If the coefficients of the quadratic equation are rational, then the irrational roots occur in conjugate pairs so if one root is α = 1/(2 + √5) = √5 – 2, then the other root will be β = 1/(2 – √5) = -√5 – 2.

Sum of the roots α + β = -4 and product of roots α β = -1.

Thus, the required equation is x^2 + 4x – 1 = 0

Form a quadratic equation with real coefficients when one of its roots is (3 – 5i).

Since the complex roots always occur in the form of pairs, the other root is 3 + 5i. Therefore, by getting the sum and the product of the quadratic roots, we can form the required the formula for quadratic equation

The sum of the roots is

=(3 + 5i) + (3 – 5i) = 6.

The product of the root is

=(3)^2 - (5i)^2

=9 - 25(-1)

=9 + 25

=34

Hence, the formula of quadratic equation is x^2 – Sx + P = 0

Therefore, x^2 – 6x + 34 = 0 is the required second degree equation.

If α and β are roots of a Quadratic Equation ax^2 + bx + c then we can say that,

- α + β = -b/a
- αβ = c/a
- α – β = ±√[(α + β)2 – 4αβ]
- |α + β| = √D/|a|

Consider the relationship between the roots and coefficient of an algebraic polynomial expression could be derived by simplifying the polynomials and substituting the above results

- α^2β + β^2α = αβ (α + β) = – bc/a2
- α^2 + αβ + β2 = (α + β)^2 – αβ = (b2 – ac)/a^2
- α^2 + β^2 = (α – β)^2 – 2αβ
- α^2 – β^2 = (α + β) (α – β)
- α^3 + β^3 = (α + β)^3 + 3αβ(α + β)
- α^3 – β^3 = (α – β)^3 + 3αβ(α – β)
- (α/β)^2 + (β/α)^2 = α4 + β4/α^2β^2

Consider β to be the common root (solution) of quadratic square equations a1x^2 + b1x+c1=0and a2x^2+ b2x+c2=0. This implies that a1β2 + b1β + c1 = 0 and a2β^2 + b2β + c2 = 0.

Right now, Solving for β^2 and β then we will obtain:

β^2/(b1c2 – b2c1) = -β/(a1c2 – a2c1) = 1/(a1b2 – a2b1) [using determinant method]

Therefore, β^2 = (b1c2 – b2c1)/ (a1b2 – a2b1) . . . . . . . . . . . . . . . . (1)

And, β = (a2c1 – a1c2)/(a1b2 – a2b1) . . . . . . . . . . . . . . . . (2)

On squaring equation (2) and equating it with equation (1) we will get:

(a1b2 – a2b1)/(b1c2 – b2c1) = (a2c1 – a1c2)^2

Hence, it is the required condition for squared equations having one common root.

If both the roots of quadratic equations a1x2 + b1x + c1 and a2x2 + b2x + c2 are common then:

a1/a2 = b1/b2 = c1/c2

If α is a repeated root in the equation, i.e., the two roots are α, α of the equation f(x) = 0, then α will be a root of the derived equation

f’(x) = 0 where f’(x) = df/dx

If α is a repeated root common in the equation f(x) = 0 and Ï•(x) = 0, then α is a common root both in f’(x) = 0 and Ï• ‘(x) = 0.

There are two methods to solve second degree equation roots

- Algebraic Method
- Graphical Method

General Form: ax^2 + bx + c = 0;

x^2 + bx/a + c/a = 0

⇒ (x + b/2a)^2 = b^2/4a^2 – c/a

Or, (x + b/2a)^2 = (b2 – 4ac)/4a^2

Or, x + b/2a = ± (√b^2 – 4ac)/2a

⇒ x = [-b ± √(b^2 – 4ac)]/2a

α+β= -b/a, α.β = c/a

Therefore, quadratic equations formula can be written as,

⇒ x^2 – (α + β)x + (α.β) = 0.

Consider a second degree equation ax^2 + bx + c = 0, where a ≠ 0 and a, b, and c are real. This quadratic equation can be further rewrite as:

a[(x + b/2a)^2 + (D/4a^2)]

The given second degree equation shows a parabola whose vertex is at P [-b/2a, -D/4a] and its axis is parallel to the y-axis.

In a quadratic equation, the value of ‘a’ can be calculated whether the graph of the equation concave upwards (a > 0) or concave downwards (a < 0). On the other hand, the value of discriminant (b^2 – 4ac) resolves whether the graph of a quadratic equation will:

- Intersect the x-axis at two points i.e. b^2 – 4ac > 0
- Just touches the x-axis i.e. b^2 – 4ac = 0
- Never intersects the x-axis i.e. b^2 – 4ac < 0

Consider a quadratic expression f(x) = ax2 + bx + c, where a, b, and c are real and a ≠ 0. The quadratic roots formula of range can be further written as f(x) = x^2 + bax = ca.

Find the range of k for which 8 lies between the roots quadratic (square) equation x^2 + 3(k – 3)x + 9 = 0.

8 will lie between the roots of the quadratic expression f(x) = x^2 + 3(k – 3)x + 9 if,

f (8) < 0

i.e. 64 + 3 (k – 3) . 8 + 9 < 0,

= 64 + 24k - 72 + 9 < 0

= 24k + 1 < 0

Therefore, the range of k ∈ (∞, 24)

Generally, the maximum and minimum value for the quadratic formula quadratic equation F(x) = ax2 + bx + c = 0 can be followed in the below graphs.

- The quadratic expression has a minimum value at x = -b/2a for positive values of a (a > 0),
- The quadratic expression has a maximum value at x = -b/2a for negative value of a (a < 0),

The maximum and minimum values of the quadratic equations are of further help to find the range of the quadratic expression:

The range of the quadratic expressions also depends on the value of a.

- For positive value of a > 0, Range: [ f(-b/2a), ∞)
- For for negative value of a < 0, Range: (-∞, f(-b/2a)]

Note that the domain of a quadratic function is the set of all real numbers in the equation, i.e., (-∞, ∞).

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