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).
Example: 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: 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.
Relationship between Roots and Coefficients
If α1,α2,α3,α4,α5,α6,............,αn are the roots of the quadratic Equation:
Example: 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: 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: 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: 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