 # 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).

Example: 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: 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.

### 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 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: 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: 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: 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: 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) = $\mathbf{\left ( \frac{1}{2}\;-\; \frac{3}{2}\right )\;=\;-\;1}$