Every Equation of nth degree has a total ‘n’ real or imaginary roots. If α is the root of Equation f (x) = 0, then the polynomial f (x) is exactly divisible by (x – α) i.e. (x – α) is the factor of the given polynomial f (x).
In algebra, the study of algebraic equations which are equations defined by a polynomial is called the theory of equations. Apolynomial is an expression consisting of one or more terms. A main difficulty of the theory of equations was to know when an algebraic equation has an algebraic solution. In this article, we will learn about the theory of equations and examples of solving equations.
Following are some important concepts covered under the theory of equations.
Simultaneous linear equations
Finding the integer solutions of an equation or of a system of equations
Systems of polynomial equations
Important Points to Remember:
The important concepts in the theory of equations are given below.
The general form of a quadratic equation in x is given by ax2 + bx + c = 0
The roots are given by x = (-b±√(b2 – 4ac))/2a
If α and β are the roots of the equation ax2 + bx + c = 0, a ≠ 0 , then sum of roots, α + β = -b/a.
Product of roots, αβ = c/a
If sum and product of roots are known, then quadratic equation is given by x2 – (sum of roots)x + product of roots = 0
For a quadratic equation, b2 – 4ac is known as the discriminant denoted by D.
If D = 0, the equation will have two equal real roots.
If D > 0, then the equation will have two distinct real roots.
If D < 0, then the equation has no real roots.
The graph of a quadratic equation is a parabola. The parabola will open upwards if a >0, open downwards if a < 0.
If a > 0, when x = -b/2a, f(x) attains its minimum value.
If a < 0, when x = -b/2a, f(x) attains its maximum value.
Relationship between Roots and Coefficients
If α1,α2,α3,α4,α5,α6,............,αn are the roots of the quadratic Equation:
Example 3: If α and β are the roots of the Quadratic Equation 2x2+6x+k=0. Find the Maximum value of [βα+αβ] if k < 0.
From the Given Quadratic Equation 2x2+6x+k=0,
α + β = – 3 and αβ = 2k
Since, k < 0, therefore, D = (b2–4ac) = 36 – 4k > 0.
Hence, the roots of quadratic equation α and β are real.
Now, [βα+αβ] = [αβα2+β2]
Since, k < 0, therefore, the maximum possible value of the above expression is -2.
Hence the maximum value of [βα+αβ] = – 2.
Example 4: In a polynomial equation px4+qx3+rx3+sx2+tx+u=0, the product of all the roots taken at a time is one third of the sum of the product of roots taken two at a time. Find the relationship between ‘r’ and ‘u’.
Let f (x) = px4+qx3+rx3+sx2+tx+u
The product of all the roots = pu
And, the sum of the product of roots taken two at a time = pr
Now, according to the given condition:
pr = 3pu
Therefore, r = 3u
Example 5: Solve the equation log4(2x2+x+1)–log2(2x–1)=1.
Given, f (x) = log4(2x2+x+1)–log2(2x–1)–1
Or, 14x2−17x+3 = 0
i.e. (x – 1) (14x – 3) = 0
Since, x = 143 does not satisfy the given equation.
Therefore, x = 1 is the solution of the given equation f (x).
Example 6: Find all the roots of equation f (x) = x4–21x2–2x3+22x+40, if its roots are in A.P.
Let a – 3d, a – d, a + d, and a + 3d be the roots of the given equation.
Now, The Sum of Roots = 4a = −ab = 2
Therefore, a = 21
And, the product of roots = (a–3d)(a–d)(a+d)(a+3d)=(a2–9d2)(a2–d2) = ae = 40
Since, a = 21
Hence, y = 49 or y = −367
Neglecting y = −367
Therefore, d2=y = 49
i.e. d = −23 and d = 23
Therefore, the roots of equation f (x):
= (a – 3d) = (21−3×23)=−4
= (a + 3d) = (21+3×23)=5
= (a + d) = (21+23)=2
= (a – d) = (21−23)=−1
Example 7: Solve the Equation 3x3–x+88=4x2 if one of the root of the given cubic polynomial equation is 2−7i.
Let, f (x) = 3x3–x+88–4x2
The given cubic function will have three roots i.e. α, β, and γ.
Therefore, α = x = 2−7i
And, β = x = 2+7i [Since, imaginary roots occur in conjugate pairs]
Therefore, (x−2+7i)(x−2−7i) = (x–2)2+7=x2–4x+11 = g (x)
On dividing f (x) by g (x) we will get 3x + 8 as quotient.
Therefore, 3x + 8 = 0
Therefore, the roots of given Equation f (x) are 2−7i, 2+7i, γ=−38
⇒ If any given equation f (x) of nth degree has maximum ‘p’ positive real roots and ‘q’ negative real roots, then the given equation has at least n – (p + q) imaginary roots.
Example 8: Find how many total positive real roots, negative real roots and imaginary roots does the equation 3x7–x5+2x3–7=0 will have.
Let, f (x) = 3x7–x5+2x3–7
Now, Sign of f (x) = +ve→−ve→+ve→−ve
Since, the sign of f (x) changes thrice. Therefore, f (x) can’t have more than 3 positive real roots. [p = 3]
Now, Sign of f (- x) = −3x7+x5–2x3–7=0 = −ve→+ve→−ve→−ve
Since, the sign of f (- x) changes twice. Therefore, f (x) can’t have more than 2 negative real roots. [q = 2]
For Imaginary Roots: n – (p + q) = 7 + (3 + 2) = 2.
Therefore, f (x) will have at least two imaginary roots.
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