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Quadratic and Cubic Equations in One Variable

Since now you are quite familiar with the term equations. You also know the various methods of solving linear equations so as to get the roots or the solution for it. But what will happen if the variable does not vary linearly? What if they vary to some degree other than one? You may have heard of equations which are quadratic and cubic in nature. Here, we are going to learn quadratic and cubic equations.

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Quadratic Equation in One Variable

An equation in which the variable varies to a degree of two is a quadratic equation. In other words, an equation in which the variable has the maximum degree of two is a quadratic one. Quadratic means related to a square.

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The Standard form of Quadratic Equation

The general form of a quadratic equation is a x2 + b x + c = 0. Here, x is a variable and a, b, and c are constants with a ≠ 0. The number of solutions that satisfy a quadratic equation will be two. A graph of a quadratic equation is a parabola.

cubic

Methods for Solving Quadratic Equations

I. Factorization Method

The main principle of this method is the principle of zero products. According to it if the product of two numbers is zero, then at least one of the factors is zero.

If α and β are the two roots of the quadratic equation, a x2 + b x + c = 0. These roots are the factors of the equation in a way such that the sum of the roots is equal to the negative of the constant  \( \frac{b}{a} \) in the equation. The constant \( \frac{c}{a} \) is equal to the product of the roots.

In this method, we need to find the factors which on being added and multiplied will result in the respective constants of the equations. The quadratic equation is thus written as, x2 + (sum of the root) x + (product of the roots) = 0.

II. Method of Perfect Square

In this method, we try to reduce the quadratic equation into a perfect square. The steps for solving the equations are

  1. If the quadratic equation is of the form a x2 + b x + c = 0. Divide both sides of the equations by a
  2. Try to make the left-hand side a complete or a perfect square by adding the value of  (\( \frac{b}{2a} \))2 in the left. Add the right-hand side by the same (\( \frac{b}{2a} \))2
  3. Take the square root of both sides
  4.  Solve the like terms in the equation to find the value(s) of x. This will give the roots of the variable

III. Using Quadratic Formula

This method is the direct application of the method of the perfect square. The roots of the equation are obtained directly using the formula

cubic

Putting the values in the above formula we find two roots, one for the positive value of the square root and one for the negative value of the square root.

Nature of the Roots

Let us denote √(b2 − 4ac) by D. Here, D is the discriminant of the equation. D determines the nature of the roots.

  • If D = 0, the roots are real and equal
  • If D < 0, then the roots are imaginary
  • The roots are real and distinct if D > 0
  • For D > 0, if D is a perfect square also then the roots are real, rational and unequal
  • If D > 0 but not a perfect square, then the roots are real, irrational and distinct

Cubic Equation in One Variable

An equation in which the variable varies to a degree of three is a cubic equation. In other words, an equation in which the variable has the maximum degree of three is a cubic one.

The General form of Cubic Equation

The general form of a cubic equation is a x3 + b x2 + cx + d = 0. Here, x is a variable and a, b, c, and d are constants with a ≠ 0. The number of solutions that satisfy a cubic equation will be three. They may or may not be equal. A graph of a cubic equation is of the form:

cubic

The Relation Between Roots and Coefficients

If α, β, and γ are the three roots of a cubic equation a x3 + b x2 + cx + d = 0.

  • Sum of the roots is equal to the negative of the constant \(\frac{b}{a}\)  in the equation
  • The product of the roots is equal to the negative of the constant \(\frac{d}{a}\)
  • The constant \(\frac{c}{a}\) is equal to the sum of the product of the roots
Root Expression Equal to
α + β + γ -\(\frac{b}{a}\)
αβ + βγ+ γα \(\frac{c}{a}\)
αβγ -\(\frac{d}{a}\)

Methods for Solving Cubic Equation

I. Factoring by Grouping

  1. The first step is to group the cubic equation into sections
  2. Next step is to find the common terms in the sections. One of the sections will have the variable and the other one will have the constant in the common
  3. Factor the common terms of the two sections
  4. Combine the factors together if each of the two terms contains the same factor
  5. Equate this with zero and find the solution of the cubic equations

II. Division Method

Suppose a cubic equation of the form a x3 + b x2 + cx + d = 0.

  • Find all the factors of d
  • Putting x equal to the various factors of d, try to make the polynomial equal zero. Let that value be k. This k is one of the roots of the cubic equation
  • Divide the cubic equation by x – k
  • Write the cubic equation as the product of the divisor (x – k) and the quotient
  • Factorize the quotient term (quadratic equation) by any of the methods. This will give the roots of the cubic equations

Solved Example for You

Problem: Solve the quadratic equation 2x2 + 4x – 1 = 0.

Solution: By using the quadratic formula, we have D = √(b2 − 4ac)= √(42 – 4 × 2 × – 1) = √(16 + 8) = √24 = 2√6

x = \( \frac{(– b ± D)} {2a}\ )\

or, x = (– 4 + 2√6) ⁄ 4 = \(\frac{1}{2}\)\ (– 2 + √6), and x = (– 4 – 2√6) / 4 = –  \(\frac{1}{2}\)\ ( 2 + √6).

So, the roots are  \(\frac{1}{2}\)\ (– 2 + √6), and  –  \(\frac{1}{2}\)\ ( 2 + √6).

Problem: Solve the cubic equation x3 – 6 x2 + 11x – 6 = 0.

Solution: Let us use the division method for solving a cubic equation. Taking the factor of d (= 6 neglecting the sign) and trying to equate the equation zero we have x = 2. Dividing the given cubic equation by x – 2, we can rewrite the equation as (x – 2) (x2 – 4x + 3) = 0. Solving the quadratic equation, x2 – 4x + 3, we have x = 1, and 3. So, the roots of the cubic equations are 1, 2, and 3.

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