# Theory of Equations

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. A polynomial 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.

• Linear 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.

1. The general form of a quadratic equation in x is given by ax2 + bx + c  = 0
2. The roots are given by x = (-b±√(b2 – 4ac))/2a
3. 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

1. If sum and product of roots are known, then quadratic equation is given by x2 – (sum of roots)x + product of roots = 0
1. For a quadratic equation, b2 – 4ac is known as the discriminant denoted by D.
2. If D = 0, the equation will have two equal real roots.
3. If D > 0, then the equation will have two distinct real roots.
4. If D < 0, then the equation has no real roots.
5. The graph of a quadratic equation is a parabola. The parabola will open upwards if a >0, open downwards if a < 0.
6. If a > 0, when x = -b/2a, f(x) attains its minimum value.
7. If a < 0, when x = -b/2a, f(x) attains its maximum value.

## Relationship between Roots and Coefficients

If $α_{1}, α_{2}, α_{3}, α_{4}, α_{5}, α_{6}, . . . . . . . . . . . . , α_{n}$ are the roots of the quadratic Equation:

$a_{0} ­x^{n} + a_{1} ­x^{n-1} + a_{2} ­x^{n – 2} + a_{3} ­x^{n – 3} + a_{4} ­x^{n – 4 }+ . . . . . . . . . . . . . . . . + a_{n-1} x + a_{n} = 0$

Then, the Sum of Roots:

$\sum \alpha_{1} = \alpha_{1} + \alpha_{2} + \alpha_{3} + \alpha_{4} . . . . . . \alpha_{n}$ = $\mathbf{-\;\frac{a_{1}}{a_{o}}}$

Sum of Product of roots taken two at a time:

$\sum \alpha_{1} \alpha_{2}$ = $\mathbf{\frac{a_{2}}{a_{o}}}$

Sum of Product of roots taken three at a time:

$\sum \alpha_{1} . \alpha_{2} .\alpha_{3}$ = $\mathbf{-\;\frac{a_{3}}{a_{o}}}$

Product of Roots:

$\alpha_{1} . \alpha_{2} . \alpha_{3} . \alpha_{4} . \alpha_{5}. . . . . . \alpha_{n} = (- 1)^{n} \times \frac{a_{n}}{a_{o}}$

Therefore, For a Cubic Equation $ax^{3} + bx^{2} + cx + d = 0$

$\alpha{1} + \alpha_{2} + \alpha_{3}$ = $\mathbf{-\;\frac{a_{1}}{a_{o}}}$ = $\mathbf{-\;\frac{b}{a}}$

$\alpha_{1}.\alpha_{2} + \alpha_{2}. \alpha_{3}+ \alpha_{3}.\alpha_{1}$ = $\mathbf{\frac{a_{2}}{a_{o}}}$ = $\mathbf{\frac{c}{a}}$

$\alpha_{1}.\alpha_{2} . \alpha_{3} = (- 1)^{n} \times \frac{a_{3}}{a_{o}}$ = $-\frac{d}{a}$

## Theory of Equation Solved Problems

Example 1: If one of the root of cubic equation is double of another, then find all the roots of Equation $x^{3} + 36 = 7x^{2}$.

Solution:

Given, f (x) = $x^{3} – 7x^{2} + 0x + 36$.

Let α, β, and γ be the roots of the given cubic function f (x).

Therefore, α + β + γ = $\mathbf{-\;\frac{b}{a}}$ = 7 . . . . . . (1)

Also, α β + γ β + αγ = $\mathbf{\frac{c}{a}}$ = 0 . . . . . . . . . . (2)

And, αβγ = $\mathbf{-\;\frac{d}{a}}$ = – 36 . . . . . . . . (3)

Since, α = 2β [Given]

Therefore, 3β + γ = 7 [From Equation (1)] . . . . . . . . . (4)

Also, 2β$^{2}$ + 3βγ = 0 [From Equation (2)]

β (2β + 3γ) = 0

Since, β ≠ 0, Therefore, 2β + 3γ = 0 . . . . . . . . . . . . . (5)

And, 2β$^{2}$ γ = – 36 [From Equation (3)] . . . . . . . . . . . . . . (6)

On Solving Equation (4) and Equation (5) we get,

β = 3 and γ = – 2

Therefore, the roots of equation $x^{3} + 36 = 7x^{2}$ are 3, 6, and – 2.

Example 2: If x, y, and z are real variables satisfying the equations x + y + z = 5 and xy + yz+ zx = 8. Determine the range of x.

Solution:

Since, x + y + z = 5 (Given)

Therefore, z = 5 – (y + x) . . . . . . . . . . . . (1)

On Substituting the values of Equation (1) in xy + yz+ zx = 8 we get,

xy + z (y+ z) = 8,

i.e. xy + (y + z) (5 – y- x) = 8,

Or, $xy + 5y – y^{2} – yx + 5x – xy – x^{2} = 8$

Or, $y^{2} – (5 – x)y – 5x + 8 + x^{2} = 0$

Now, $b^{2} – 4ac (D) ≥ 0$ [Since, y is real]

i.e. $(5 – x)^{2} – 4 (x^{2} – 5x + 8) ≥ 0$,

Or, $x^{2} + 25 – 10x – 4x^{2} + 20x – 32 ≥ 0$,

Or,$- 3x^{2} + 10x – 7 ≥ 0$,

Or, $3x^{2} – 10x + 7 ≤ 0$ . . . . . . . . . . . (2)

Now, the roots of quadratic equation $3x^{2} – 10x + 7 = 0$

x = $\mathbf{\frac{+\;10\;\pm \;\sqrt{100\;-\;84}}{6}\;=\;\frac{10\;\pm \;4}{6}\;=\;\frac{7}{3},\;1}$

Therefore, From Equation (2),

$\mathbf{\left ( x\;-\;1 \right )\;\left ( x\;-\;\frac{7}{3} \right )}$ ≤ 0

i.e. $\mathbf{x\;\in \;\left [ 1, \;\frac{7}{3} \right ]}$

Example 3: If α and β are the roots of the Quadratic Equation $2x^{2} + 6x + k = 0$. Find the Maximum value of $\mathbf{\left [\;\frac{\alpha }{\beta }\;+\;\frac{\beta }{\alpha }\; \right ]}$ if k < 0.

Solution:

From the Given Quadratic Equation $2x^{2} + 6x + k = 0$,

α + β = – 3 and αβ = $\mathbf{\frac{k}{2}}$

Since, k < 0, therefore, D = ($b^{2} – 4ac$) = 36 – 4k > 0.

Hence, the roots of quadratic equation α and β are real.

Now, $\mathbf{\left [\;\frac{\alpha }{\beta }\;+\;\frac{\beta }{\alpha }\; \right ]}$ = $\mathbf{\left [\;\frac{\alpha^{2} \;+\;\beta ^{2}}{\alpha \;\beta }\right ]}$

= $\mathbf{\frac{\left ( \alpha \;+\; \beta \right )^{2}\;-\;2\;\alpha \;\beta}{\alpha \;\beta }\;=\; \frac{\left ( \alpha \;+\; \beta \right )^{2}}{\alpha\; \beta }\;-\;2}$

= $\mathbf{\frac{18}{k}\;-\;2}$

Since, k < 0, therefore, the maximum possible value of the above expression is -2.

Hence the maximum value of $\mathbf{\left [\;\frac{\alpha }{\beta }\;+\;\frac{\beta }{\alpha }\; \right ]}$ = – 2.

Example 4: In a polynomial equation $px^{4} + qx^{3} + rx^{3} + sx^{2} + tx + u = 0$, 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’.

Solution:

Let f (x) = $px^{4} + qx^{3} + rx^{3} + sx^{2} + tx + u$

The product of all the roots = $\mathbf{\frac{u}{p}}$

And, the sum of the product of roots taken two at a time = $\mathbf{\frac{r}{p}}$

Now, according to the given condition:

$\mathbf{\frac{r}{p}}$ = $3\;\mathbf{\frac{u}{p}}$

Therefore, r = 3u

Example 5: Solve the equation $log_{4} (2x^{2} + x + 1) – log_{2} (2x – 1) = 1$.

Solution:

Given, f (x) = $log_{4} (2x^{2} + x + 1) – log_{2} (2x – 1) – 1$

i.e. $\mathbf{\frac{log_{e}\;(2\;x^{2}\;+\;x\;+\;1)}{log_{e}\;4}\;-\;\frac{log_{e}\;(2x\;-\;1)}{log_{e}\;2}\;=\;1}$

Or, $\mathbf{\frac{log_{e}\;(2\;x^{2}\;+\;x\;+\;1)}{log_{e}\;4}\;-\;\frac{log_{e}\;(2x\;-\;1)^{2}}{log_{e}\;4}\;=\;1}$

Or, $\mathbf{log_{e}\;\frac{(2\;x^{2}\;+\;x\;+\;1)}{(2x\;-\;1)^{2}}\;=\;log_{e}\;4}$

Or, $\mathbf{2\;x^{2}\;+\;x\;+\;1 \;=\; 16\;x^{2}\;+\;4\;-\;16x}$

Or, $\mathbf{14\;x^{2}\;-\;17\;x\;+\;3}$ = 0

i.e. (x – 1) (14x – 3) = 0

Since, x = $\mathbf{\frac{3}{14}}$ 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) = $x^{4} – 21 x^{2} – 2 x^{3} + 22x + 40$, if its roots are in A.P.

Solution:

Let a – 3d, a – d, a + d, and a + 3d be the roots of the given equation.

Now, The Sum of Roots = 4a = $\mathbf{-\;\frac{b}{a}}$ = 2

Therefore, a = $\mathbf{\frac{1}{2}}$

And, the product of roots = $(a – 3d) (a – d) (a + d) (a + 3d) = (a^{2} – 9d^{2}) (a^{2} – d^{2})$ = $\mathbf{\frac{e}{a}}$ = 40

Since, a = $\mathbf{\frac{1}{2}}$

Therefore, $\mathbf{\left ( \frac{1}{4}\;-\;9\;d^{2} \right )\;\left ( \frac{1}{4}\;-\;d^{2} \right )\;=\;40}$

Or, $144d^{4} – 40d^{2} – 639 = 0$,

Let, $d^{2} = y$

Therefore, $144y^{2} – 40y^{2} – 639 = 0$

Hence, y = $\mathbf{\frac{9}{4}}$ or y = $\mathbf{-\;\frac{7}{36}}$

Neglecting y = $\mathbf{-\;\frac{7}{36}}$

Therefore, $d^{2} = y$ = $\mathbf{\frac{9}{4}}$

i.e. d = $\mathbf{-\;\frac{3}{2}}$ and d = $\mathbf{\frac{3}{2}}$

Therefore, the roots of equation f (x):

= (a – 3d) = $\mathbf{\left ( \frac{1}{2}\;-\;3\times \frac{3}{2}\right )\;=\;-\;4}$

= (a + 3d) = $\mathbf{\left ( \frac{1}{2}\;+\;3\times \frac{3}{2}\right )\;=\;5}$

= (a + d) = $\mathbf{\left ( \frac{1}{2}\;+\; \frac{3}{2}\right )\;=\;2}$

= (a – d) = $\mathbf{\left ( \frac{1}{2}\;-\; \frac{3}{2}\right )\;=\;-\;1}$

Example 7: Solve the Equation $3x^{3} – x + 88 = 4x^{2}$ if one of the root of the given cubic polynomial equation is $\mathbf{2\;-\;\sqrt{7}\;i}$.

Solution:

Let, f (x) = $3x^{3} – x + 88 – 4x^{2}$

The given cubic function will have three roots i.e. α, β, and γ.

Therefore, α = x = $\mathbf{2\;-\;\sqrt{7}\;i}$

And, β = x = $\mathbf{2\;+\;\sqrt{7}\;i}$ [Since, imaginary roots occur in conjugate pairs]

Therefore, $\mathbf{(x\;-\;2\;+\;\sqrt{7}\;i)\;(x\;-\;2\;-\;\sqrt{7}\;i)}$ = $(x – 2)^{2} + 7 = x^{2} – 4x + 11$ = g (x)

On dividing f (x) by g (x) we will get 3x + 8 as quotient.

Therefore, 3x + 8 = 0

Hence, $\mathbf{\gamma \;=\;-\;\frac{8}{3}}$

Therefore, the roots of given Equation f (x) are $\mathbf{2\;-\;\sqrt{7}\;i}$, $\mathbf{2\;+\;\sqrt{7}\;i}$, $\mathbf{\gamma \;=\;-\;\frac{8}{3}}$

⇒ 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 $3x^{7} – x^{5} + 2x^{3} – 7 = 0$ will have.

Solution:

Let, f (x) = $3x^{7} – x^{5} + 2x^{3} – 7$

Now, Sign of f (x) = $\mathbf{+\;ve\;\rightarrow \;-\;ve\;\rightarrow \;+\;ve\;\rightarrow \;-\;ve}$

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) = $-3x^{7} + x^{5} – 2x^{3} – 7 = 0$ = $\mathbf{-\;ve\;\rightarrow \;+\;ve\;\rightarrow \;-\;ve\;\rightarrow \;-\;ve}$

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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